Q1. Evaluate the following using direct method:
(i) $(x + 8)(x + 3)$
(ii) $(y + 5)(y -3)$
(iii) $(a - 8)(a + 2)$
(iv) $(b - 3)(b - 5)$
(v) $(3x - 2y)(2x + y)$
(vi) $(5a + 16)(3a - 7)$
(vii) $(8 - b)(3 + b)$
Show Answer
(i) $(x + 8)(x + 3)$
$\Rightarrow x^2 + (8x + 3x) + 24$
$\Rightarrow x^2 + 8x + 3x + 24$
$\Rightarrow x^2 + 11x + 24$ Ans
(ii) $(y + 5)(y - 3)$
$\Rightarrow y^2 + (-3y + 5y) + (-15)$
$\Rightarrow y^2 + 2y - 15$ Ans
(iii) $(a - 8)(a + 2)$
$\Rightarrow a^2 + (2a - 8a) + (-16)$
$\Rightarrow a^2 + (-6a) - 16$
$\Rightarrow a^2 - 6a - 16$ Ans
(iv) $(b - 3)(b - 5)$
$\Rightarrow b^2 + (-5b - 3b) + 15$
$\Rightarrow b^2 + (-8b) + 15$
$\Rightarrow b^2 - 8b + 15$ Ans
(v) $(3x - 2y)(2x + y)$
$\Rightarrow 6x^2 + (3xy - 4xy) + (-2y^2)$
$\Rightarrow 6x^2 + (-xy) - 2y^2$
$\Rightarrow 6x^2 - xy - 2y^2$ Ans
(vi) $(5a + 16)(3a - 7)$
$\Rightarrow 15a^2 + (-35a + 48a) + (-112)$
$\Rightarrow 15a^2 + 13a - 112$ Ans
(vii) $(8 - b)(3 + b)$
$\Rightarrow 24 + (8b - 3b) +(-b^2)$
$\Rightarrow 24 + 5b - b^2$ Ans
$\Rightarrow x^2 + (8x + 3x) + 24$
$\Rightarrow x^2 + 8x + 3x + 24$
$\Rightarrow x^2 + 11x + 24$ Ans
(ii) $(y + 5)(y - 3)$
$\Rightarrow y^2 + (-3y + 5y) + (-15)$
$\Rightarrow y^2 + 2y - 15$ Ans
(iii) $(a - 8)(a + 2)$
$\Rightarrow a^2 + (2a - 8a) + (-16)$
$\Rightarrow a^2 + (-6a) - 16$
$\Rightarrow a^2 - 6a - 16$ Ans
(iv) $(b - 3)(b - 5)$
$\Rightarrow b^2 + (-5b - 3b) + 15$
$\Rightarrow b^2 + (-8b) + 15$
$\Rightarrow b^2 - 8b + 15$ Ans
(v) $(3x - 2y)(2x + y)$
$\Rightarrow 6x^2 + (3xy - 4xy) + (-2y^2)$
$\Rightarrow 6x^2 + (-xy) - 2y^2$
$\Rightarrow 6x^2 - xy - 2y^2$ Ans
(vi) $(5a + 16)(3a - 7)$
$\Rightarrow 15a^2 + (-35a + 48a) + (-112)$
$\Rightarrow 15a^2 + 13a - 112$ Ans
(vii) $(8 - b)(3 + b)$
$\Rightarrow 24 + (8b - 3b) +(-b^2)$
$\Rightarrow 24 + 5b - b^2$ Ans
Q2. Evaluate the following using direct method:
(i) $(x + 1)(x - 1)$
(ii) $(2 + a)(2 - a)$
(iii) $(3b - 1)(3b + 1)$
(iv) $(4 + 5x)(4 - 5x)$
(v) $(2a + 3)(2a - 3)$
(vi) $(xy + 4)(xy - 4)$
(vii) $(ab + x^2)(ab - x^2)$
(viii) $(3x^2 + 5y^2)(3x^2 - 5y^2)$
(ix) $(z - \dfrac{2}{3})(z + \dfrac{2}{3})$
(x) $(\dfrac{3}{5}a + \dfrac{1}{2})(\dfrac{3}{5}a - \dfrac{1}{2})$
(xi) $(0.5 - 2a)(0.5 + 2a)$
(xii) $(\dfrac{a}{2} - \dfrac{b}{3})(\dfrac{a}{2} + \dfrac{b}{3})$
Show Answer
(i) $(x + 1)(x - 1)$
$\Rightarrow x^2 + (-x + x) + (-1)$
$\Rightarrow x^2 - 1$ Ans
(ii) $(2 + a)(2 - a)$
$\Rightarrow 4 + (-2a + 2a) + (-a^2)$
$\Rightarrow 4 - a^2$ Ans
(iii) $(3b - 1)(3b + 1)$
$\Rightarrow 9b^2 + (3b - 3b) + (-1)$
$\Rightarrow 9b^2 - 1$ Ans
(iv) $(4 + 5x)(4 - 5x)$
$\Rightarrow 16 + (-20x + 20x) + (-25x^2)$
$\Rightarrow 16 - 25x^2$ Ans
(v) $(2a + 3)(2a - 3)$
$\Rightarrow 4a^2 + (-6a + 6a) + (-9)$
$\Rightarrow 4a^2 - 9$ Ans
(vi) $(xy + 4)(xy - 4)$
$\Rightarrow (xy)^2 + (-4xy + 4xy) + (-16)$
$\Rightarrow x^2y^2 - 16$ Ans
(vii) $(ab + x^2)(ab - x^2)$
$\Rightarrow (ab)^2 + (-abx^2 + abx^2) + (-x^2)^2$
$\Rightarrow a^2b^2 - x^4$ Ans
(viii) $(3x^2 + 5y^2)(3x^2 - 5y^2)$
$\Rightarrow (3x^2)^2 + (-15x^2y^2 + 15x^2y^2) + (-5y^2)^2$
$\Rightarrow 9^4 - 25^4$ Ans
(ix) $(z - \dfrac{2}{3})(z + \dfrac{2}{3})$
$\Rightarrow z^2 + (\dfrac{2}{3}z - \dfrac{2}{3}z) + (-\dfrac{2}{3})^2$
$\Rightarrow z^2 - \dfrac{4}{9}$ Ans
(x) $(\dfrac{3}{5}a + \dfrac{1}{2})(\dfrac{3}{5}a - \dfrac{1}{2})$
$(\dfrac{3}{5}a)^2 + (-\dfrac{3}{10}a + \dfrac{3}{10}a) + (-\dfrac{1}{2})^2$
$\dfrac{9}{25}a^2 - \dfrac{1}{4}$ Ans
(xi) $(0.5 - 2a)(0.5 + 2a)$
$\Rightarrow (0.5)^2 + (1a - 1a) + (-2a)^2$
$\Rightarrow 0.25 -4a^2$ Ans
(xii) $(\dfrac{a}{2} - \dfrac{b}{3})(\dfrac{a}{2} + \dfrac{b}{3})$
$\Rightarrow (\dfrac{a}{2})^2 + (\dfrac{ab}{6} - \dfrac{ab}{6}) + (-\dfrac{b}{3})^2$
$\Rightarrow \dfrac{a^2}{4} - \dfrac{b^2}{9}$ Ans
$\Rightarrow x^2 + (-x + x) + (-1)$
$\Rightarrow x^2 - 1$ Ans
(ii) $(2 + a)(2 - a)$
$\Rightarrow 4 + (-2a + 2a) + (-a^2)$
$\Rightarrow 4 - a^2$ Ans
(iii) $(3b - 1)(3b + 1)$
$\Rightarrow 9b^2 + (3b - 3b) + (-1)$
$\Rightarrow 9b^2 - 1$ Ans
(iv) $(4 + 5x)(4 - 5x)$
$\Rightarrow 16 + (-20x + 20x) + (-25x^2)$
$\Rightarrow 16 - 25x^2$ Ans
(v) $(2a + 3)(2a - 3)$
$\Rightarrow 4a^2 + (-6a + 6a) + (-9)$
$\Rightarrow 4a^2 - 9$ Ans
(vi) $(xy + 4)(xy - 4)$
$\Rightarrow (xy)^2 + (-4xy + 4xy) + (-16)$
$\Rightarrow x^2y^2 - 16$ Ans
(vii) $(ab + x^2)(ab - x^2)$
$\Rightarrow (ab)^2 + (-abx^2 + abx^2) + (-x^2)^2$
$\Rightarrow a^2b^2 - x^4$ Ans
(viii) $(3x^2 + 5y^2)(3x^2 - 5y^2)$
$\Rightarrow (3x^2)^2 + (-15x^2y^2 + 15x^2y^2) + (-5y^2)^2$
$\Rightarrow 9^4 - 25^4$ Ans
(ix) $(z - \dfrac{2}{3})(z + \dfrac{2}{3})$
$\Rightarrow z^2 + (\dfrac{2}{3}z - \dfrac{2}{3}z) + (-\dfrac{2}{3})^2$
$\Rightarrow z^2 - \dfrac{4}{9}$ Ans
(x) $(\dfrac{3}{5}a + \dfrac{1}{2})(\dfrac{3}{5}a - \dfrac{1}{2})$
$(\dfrac{3}{5}a)^2 + (-\dfrac{3}{10}a + \dfrac{3}{10}a) + (-\dfrac{1}{2})^2$
$\dfrac{9}{25}a^2 - \dfrac{1}{4}$ Ans
(xi) $(0.5 - 2a)(0.5 + 2a)$
$\Rightarrow (0.5)^2 + (1a - 1a) + (-2a)^2$
$\Rightarrow 0.25 -4a^2$ Ans
(xii) $(\dfrac{a}{2} - \dfrac{b}{3})(\dfrac{a}{2} + \dfrac{b}{3})$
$\Rightarrow (\dfrac{a}{2})^2 + (\dfrac{ab}{6} - \dfrac{ab}{6}) + (-\dfrac{b}{3})^2$
$\Rightarrow \dfrac{a^2}{4} - \dfrac{b^2}{9}$ Ans
Q3. Evaluate the following:
(i) $(a + 1)(a - 1)(a^2 + 1)$
(ii) $(a + b)(a - b)(a^2 + b^2)$
(iii) $(2a - b)(2a + b)(4a^2 + b^2)$
(iv) $(3 - 2x)(3 + 2x)(9 + 4x^2)$
(v) $(3x - 4y)(3x + 4y)(9x^2 + 16y^2)$
Show Answer
(i) $(a + 1)(a - 1)(a^2 + 1)$
$\Rightarrow (a^2) + (-a + a) + (-1) (a^2 + 1)$
$\Rightarrow (a^2 - 1) (a^2 + 1)$
$\Rightarrow (a^2)^2 + (a^2 - a^2) + (-1)$
$\Rightarrow a^4 - 1$ Ans
(ii) $(a + b)(a - b) (a^2 + b^2)$
$\Rightarrow (a)^2 + (-ab + ab) + (-b)^2 (a^2 + b^2)$
$\Rightarrow (a^2 - b^2) (a^2 + b^2)$
$\Rightarrow (a^2)^2 + (a^2b^2 - a^2b^2) + (-b^2)^2$
$\Rightarrow a^4 - b^4$ Ans
(iii) $(2a - b)(2a + b) (4a^2 + b^2)$
$\Rightarrow (2a)^2 + (2ab - 2ab) + (-b)^2 (4a^2 + b^2)$
$\Rightarrow (4a^2 - b^2) (4a^2 + b^2)$
$\Rightarrow (4a^2 \times 4a^2) + (4a^2b^2 - 4a^2b^2) + (-b^2)^2$
$\Rightarrow 16a^4 - b^4$ Ans
(iv) $(3 - 2x)(3 + 2x)(9 + 4x^2)$
$\Rightarrow (3)^2 + (6x - 6x) + (-2x)^2 (9 + 4x^2)$
$\Rightarrow (9 - 4x^2) (9 + 4x^2)$
$\Rightarrow (9)^2 + (36x^2 - 36x^2) + (-4x^2)^2$
$\Rightarrow 81 - 16x^4$ Ans
(v) $(3x - 4y)(3x + 4y)(9x^2 + 16y^2)$
$\Rightarrow (3x)^2 + (12xy - 12xy) + (-4y)^2 (9x^2 + 16y^2)$
$\Rightarrow (9x^2 - 16y^2) (9x^2 + 16y^2)$
$\Rightarrow (9x^2)^2 + (132x^2y^2 - 132x^2y^2) + (-16y^2)^2$
$\Rightarrow 81x^4 - 256y^4$ Ans
$\Rightarrow (a^2) + (-a + a) + (-1) (a^2 + 1)$
$\Rightarrow (a^2 - 1) (a^2 + 1)$
$\Rightarrow (a^2)^2 + (a^2 - a^2) + (-1)$
$\Rightarrow a^4 - 1$ Ans
(ii) $(a + b)(a - b) (a^2 + b^2)$
$\Rightarrow (a)^2 + (-ab + ab) + (-b)^2 (a^2 + b^2)$
$\Rightarrow (a^2 - b^2) (a^2 + b^2)$
$\Rightarrow (a^2)^2 + (a^2b^2 - a^2b^2) + (-b^2)^2$
$\Rightarrow a^4 - b^4$ Ans
(iii) $(2a - b)(2a + b) (4a^2 + b^2)$
$\Rightarrow (2a)^2 + (2ab - 2ab) + (-b)^2 (4a^2 + b^2)$
$\Rightarrow (4a^2 - b^2) (4a^2 + b^2)$
$\Rightarrow (4a^2 \times 4a^2) + (4a^2b^2 - 4a^2b^2) + (-b^2)^2$
$\Rightarrow 16a^4 - b^4$ Ans
(iv) $(3 - 2x)(3 + 2x)(9 + 4x^2)$
$\Rightarrow (3)^2 + (6x - 6x) + (-2x)^2 (9 + 4x^2)$
$\Rightarrow (9 - 4x^2) (9 + 4x^2)$
$\Rightarrow (9)^2 + (36x^2 - 36x^2) + (-4x^2)^2$
$\Rightarrow 81 - 16x^4$ Ans
(v) $(3x - 4y)(3x + 4y)(9x^2 + 16y^2)$
$\Rightarrow (3x)^2 + (12xy - 12xy) + (-4y)^2 (9x^2 + 16y^2)$
$\Rightarrow (9x^2 - 16y^2) (9x^2 + 16y^2)$
$\Rightarrow (9x^2)^2 + (132x^2y^2 - 132x^2y^2) + (-16y^2)^2$
$\Rightarrow 81x^4 - 256y^4$ Ans
Q4. Use the product $(a + b)(a - b) = a^2 - b^2$ to evaluate the following:
(i) $21 \times 19$
(ii)$33 \times 27$
(iii) $103 \times 97$
(iv) $9.8 \times 10.2$
(v) $7.7 \times 8.3$
(vi) $4.6 \times 5.4$
Show Answer
(i) $21 \times 19$
$\Rightarrow (20 + 1) (20 - 1)$
$\Rightarrow (20 - 1)^2$
$\Rightarrow (20)^2 - (1)^2$
$\Rightarrow 400 - 1 = 399$ Ans
(ii)$33 \times 27$
$\Rightarrow (30 + 3) (30 - 3)$
$\Rightarrow (30 - 3)^2$
$\Rightarrow (30)^2 - (3)^2$
$\Rightarrow 900 - 9 = 891$ Ans
(iii) $103 \times 97$
$\Rightarrow (100 + 3) (100 - 3)$
$\Rightarrow (100 - 3)^2$
$\Rightarrow (100)^2 - (3)^2$
$\Rightarrow 10000 - 9 = 9991$ Ans
(iv) $9.8 \times 10.2$
$\Rightarrow (10 + 0.2) (10 - 0.2)$
$\Rightarrow (10 - 0.2)^2$
$\Rightarrow (10)^2 - (0.2)^2$
$\Rightarrow 100 - 0.04 = 99.96$ Ans
(v) $7.7 \times 8.3$
$8.3 \times 7.7$ Rearranging the order
$\Rightarrow (8 + 0.3) (8 - 0.3)$
$\Rightarrow (8 - 0.3)^2$
$\Rightarrow (8)^2 - (0.3)^2$
$\Rightarrow 64 - 0.09 = 63.91$ Ans
(vi) $4.6 \times 5.4$
$5.4 \times 4.6$ Rearranging the order
$\Rightarrow (5 + 0.4) (5 - 0.4)$
$\Rightarrow (5 - 0.4)^2$
$\Rightarrow (5)^2 - (0.4)^2$
$\Rightarrow 25 - 0.16 = 24.84$ Ans
$\Rightarrow (20 + 1) (20 - 1)$
$\Rightarrow (20 - 1)^2$
$\Rightarrow (20)^2 - (1)^2$
$\Rightarrow 400 - 1 = 399$ Ans
(ii)$33 \times 27$
$\Rightarrow (30 + 3) (30 - 3)$
$\Rightarrow (30 - 3)^2$
$\Rightarrow (30)^2 - (3)^2$
$\Rightarrow 900 - 9 = 891$ Ans
(iii) $103 \times 97$
$\Rightarrow (100 + 3) (100 - 3)$
$\Rightarrow (100 - 3)^2$
$\Rightarrow (100)^2 - (3)^2$
$\Rightarrow 10000 - 9 = 9991$ Ans
(iv) $9.8 \times 10.2$
$\Rightarrow (10 + 0.2) (10 - 0.2)$
$\Rightarrow (10 - 0.2)^2$
$\Rightarrow (10)^2 - (0.2)^2$
$\Rightarrow 100 - 0.04 = 99.96$ Ans
(v) $7.7 \times 8.3$
$8.3 \times 7.7$ Rearranging the order
$\Rightarrow (8 + 0.3) (8 - 0.3)$
$\Rightarrow (8 - 0.3)^2$
$\Rightarrow (8)^2 - (0.3)^2$
$\Rightarrow 64 - 0.09 = 63.91$ Ans
(vi) $4.6 \times 5.4$
$5.4 \times 4.6$ Rearranging the order
$\Rightarrow (5 + 0.4) (5 - 0.4)$
$\Rightarrow (5 - 0.4)^2$
$\Rightarrow (5)^2 - (0.4)^2$
$\Rightarrow 25 - 0.16 = 24.84$ Ans
Q5. Evaluate the following:
(i) $(6 - xy)(6 + xy)$
(ii) $(7x + \dfrac{2}{3}y)(7x - \dfrac{2}{3}y)$
(iii) $(\dfrac{a}{2b} + \dfrac{2b}{a})(\dfrac{a}{2b} - \dfrac{2b}{a})$
(iv) $(3x - \dfrac{1}{2y})(3x + \dfrac{1}{2y})$
(v) $(2a + 3)(2a - 3)(4a^2 + 9)$
(vi) $(a + bc)(a - bc)(a^2 + b^2c^2)$
(vii) $(5x + 8y)(3x + 5y)$
(viii) $(7x + 15y)(5x - 4y)$
(ix) $(2a - 3b)(3a + 4b)$
(x) $(9a - 7b)(3a - b)$
Show Answer
(i) $(6 - xy)(6 + xy)$
$\Rightarrow (6 - xy)^2$
$\Rightarrow (6)^2 - (xy)^2$
$\Rightarrow 36 - x^2y^2$ Ans
(ii) $(7x + \dfrac{2}{3}y)(7x - \dfrac{2}{3}y)$
$\Rightarrow (7x - \dfrac{2}{3}y)^2$
$\Rightarrow (7x)^2 - (\dfrac{2}{3}y)^2$
$\Rightarrow 49x^2 - \dfrac{4}{9}y^2$ Ans
(iii) $(\dfrac{a}{2b} + \dfrac{2b}{a})(\dfrac{a}{2b} - \dfrac{2b}{a})$
$\Rightarrow (\dfrac{a}{2b} - \dfrac{2b}{a})^2$
$\Rightarrow (\dfrac{a}{2b})^2 - (\dfrac{2b}{a})^2$
$\Rightarrow \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a^2}$ Ans
(iv) $(3x - \dfrac{1}{2y})(3x + \dfrac{1}{2y})$
$\Rightarrow (3x - \dfrac{1}{2y})^2$
$\Rightarrow (3x)^2 - (\dfrac{1}{2y})^2$
$\Rightarrow 9x^2 - \dfrac{1}{4y^2}$ Ans
(v) $(2a + 3)(2a - 3)(4a^2 + 9)$
$\Rightarrow (2a - 3)^2 (4a^2 + 9)$
$\Rightarrow (2a)^2 - (3)^2 (4a^2 + 9)$
$\Rightarrow (4a^2 - 9) (4a^2 + 9)$
$\Rightarrow (4a^2 - 9)^2$
$\Rightarrow (4a^2)^2 - (9)^2$
$\Rightarrow 16a^4 - 81$ Ans
(vi) $(a + bc)(a - bc)(a^2 + b^2c^2)$
$\Rightarrow (a - bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a)^2 - (bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2) (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2)^2$
$\Rightarrow (a^2)^2 - (b^2c^2)^2$
$\Rightarrow a^4 - b^4c^4$ Ans
(vii) $(5x + 8y)(3x + 5y)$
$\Rightarrow 3x(5x) + (25xy + 24xy) + 8y(5y)$
$\Rightarrow 15x^2 + 49xy + 40y^2$ Ans
(viii) $(7x + 15y)(5x - 4y)$
$\Rightarrow 5x(7x) + (-7x(4y) + 15y(5x)) + 15y(-4y)$
$\Rightarrow 35x^2 + (-28xy + 75xy) + (-60y^2)$
$\Rightarrow 35x^2 + 47xy - 60y^2$ Ans
(ix) $(2a - 3b)(3a + 4b)$
$\Rightarrow 3a(2a) + (2a(4b) - 3b(3a)) - 3b(4b)$
$\Rightarrow 6a^2 + (8ab - 9ab) - 12b^2$
$\Rightarrow 6a^2 + (-ab) - 12b^2$
$\Rightarrow 6a^2 - ab - 12b^2$ Ans
(x) $(9a - 7b)(3a - b)$
$\Rightarrow 3a(9a) + (-b(9a) - 7b(3a)) - 7b(-b)$
$\Rightarrow 27a^2 + (-9ab - 21ab) + 7b^2$
$\Rightarrow 27a^2 + (-30ab) + 7b^2$
$\Rightarrow 27a^2 - 30ab + 7b^2$ Ans
$\Rightarrow (6 - xy)^2$
$\Rightarrow (6)^2 - (xy)^2$
$\Rightarrow 36 - x^2y^2$ Ans
(ii) $(7x + \dfrac{2}{3}y)(7x - \dfrac{2}{3}y)$
$\Rightarrow (7x - \dfrac{2}{3}y)^2$
$\Rightarrow (7x)^2 - (\dfrac{2}{3}y)^2$
$\Rightarrow 49x^2 - \dfrac{4}{9}y^2$ Ans
(iii) $(\dfrac{a}{2b} + \dfrac{2b}{a})(\dfrac{a}{2b} - \dfrac{2b}{a})$
$\Rightarrow (\dfrac{a}{2b} - \dfrac{2b}{a})^2$
$\Rightarrow (\dfrac{a}{2b})^2 - (\dfrac{2b}{a})^2$
$\Rightarrow \dfrac{a^2}{4b^2} - \dfrac{4b^2}{a^2}$ Ans
(iv) $(3x - \dfrac{1}{2y})(3x + \dfrac{1}{2y})$
$\Rightarrow (3x - \dfrac{1}{2y})^2$
$\Rightarrow (3x)^2 - (\dfrac{1}{2y})^2$
$\Rightarrow 9x^2 - \dfrac{1}{4y^2}$ Ans
(v) $(2a + 3)(2a - 3)(4a^2 + 9)$
$\Rightarrow (2a - 3)^2 (4a^2 + 9)$
$\Rightarrow (2a)^2 - (3)^2 (4a^2 + 9)$
$\Rightarrow (4a^2 - 9) (4a^2 + 9)$
$\Rightarrow (4a^2 - 9)^2$
$\Rightarrow (4a^2)^2 - (9)^2$
$\Rightarrow 16a^4 - 81$ Ans
(vi) $(a + bc)(a - bc)(a^2 + b^2c^2)$
$\Rightarrow (a - bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a)^2 - (bc)^2 (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2) (a^2 + b^2c^2)$
$\Rightarrow (a^2 - b^2c^2)^2$
$\Rightarrow (a^2)^2 - (b^2c^2)^2$
$\Rightarrow a^4 - b^4c^4$ Ans
(vii) $(5x + 8y)(3x + 5y)$
$\Rightarrow 3x(5x) + (25xy + 24xy) + 8y(5y)$
$\Rightarrow 15x^2 + 49xy + 40y^2$ Ans
(viii) $(7x + 15y)(5x - 4y)$
$\Rightarrow 5x(7x) + (-7x(4y) + 15y(5x)) + 15y(-4y)$
$\Rightarrow 35x^2 + (-28xy + 75xy) + (-60y^2)$
$\Rightarrow 35x^2 + 47xy - 60y^2$ Ans
(ix) $(2a - 3b)(3a + 4b)$
$\Rightarrow 3a(2a) + (2a(4b) - 3b(3a)) - 3b(4b)$
$\Rightarrow 6a^2 + (8ab - 9ab) - 12b^2$
$\Rightarrow 6a^2 + (-ab) - 12b^2$
$\Rightarrow 6a^2 - ab - 12b^2$ Ans
(x) $(9a - 7b)(3a - b)$
$\Rightarrow 3a(9a) + (-b(9a) - 7b(3a)) - 7b(-b)$
$\Rightarrow 27a^2 + (-9ab - 21ab) + 7b^2$
$\Rightarrow 27a^2 + (-30ab) + 7b^2$
$\Rightarrow 27a^2 - 30ab + 7b^2$ Ans
Q6. Expand:
(i) $(2a + b)^2$
(ii) $(a - 2b)^2$
(iii) $(a + \dfrac{1}{2a})^2$
(iv) $(2a - \dfrac{1}{a})^2$
(v) $(a + b - c)^2$
(vi) $(a - b + c)^2$
(vii) $(3x + \dfrac{1}{3x})^2$
(viii) $(2x - \dfrac{1}{2x})^2$
Show Answer
(i) $(2a + b)^2$
$\Rightarrow (2a + b) (2a + b)$
$\Rightarrow (2a)^2 + (2ab + 2ab) + (b)^2$
$\Rightarrow 4a^2 + 4ab + b^2$ Ans
(ii) $(a - 2b)^2$
$\Rightarrow (a - 2b) (a - 2b)$
$\Rightarrow (a)^2 + (-2ab - 2ab) + (2b)^2$
$\Rightarrow a^2 + (-4ab) + 2b^2$
$\Rightarrow a^2 - 4ab + 2b^2$ Ans
(iii) $(a + \dfrac{1}{2a})^2$
$\Rightarrow (a + \dfrac{1}{2a}) (a + \dfrac{1}{2a})$
$\Rightarrow (a)^2 + ((\dfrac{1}{2a})a + (\dfrac{1}{2a})a) + (\dfrac{1}{2a})^2$
$\Rightarrow a^2 + (\dfrac{1}{2} + \dfrac{1}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + (\dfrac{1 + 1}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + (\dfrac{2}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + 1 + \dfrac{1}{4a^2}$ Ans
The same problem can also be solved in the following manner:
$\Rightarrow (a + \dfrac{1}{2a}) (a + \dfrac{1}{2a})$
$\Rightarrow (a)^2 + (2 \times a \times (\dfrac{1}{2a})) + (\dfrac{1}{2a})^2$
$\Rightarrow a^2 + (\dfrac{2}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + 1 + \dfrac{1}{4a^2}$ Ans
(iv) $(2a - \dfrac{1}{a})^2$
$\Rightarrow (2a - \dfrac{1}{a}) (2a - \dfrac{1}{a})$
$\Rightarrow (2a)^2 + (2 \times 2a \times (-\dfrac{1}{a})) + (\dfrac{1}{a})^2$
$\Rightarrow 4a^2 + (-\dfrac{4a}{a}) + \dfrac{1}{a^2}$
$\Rightarrow 4a^2 - 4 + \dfrac{1}{a^2}$ Ans
(v) $(a + b - c)^2$
$\Rightarrow (a + b - c) (a + b - c)$
$\Rightarrow (a^2 + ab + (-ac)) + (ab + b^2 + (-bc)) + ((-ac) + (-bc) + c^2)$
$\Rightarrow a^2 + ab - ac + ab + b^2 - bc - ac - bc + c^2$
$\Rightarrow a^2 + b^2 + c^2 + (ab + ab) + (-bc - bc) + (-ac - ac)$
$\Rightarrow a^2 + b^2 + c^2 + (2ab) + (-2bc) + (-2ac)$
$\Rightarrow a^2 + b^2 + c^2 + (2ab - 2bc - 2ac)$
$\Rightarrow a^2 + b^2 + c^2 + 2(ab - bc - ac)$ Ans
(vi) $(a - b + c)^2$
$\Rightarrow (a - b + c) (a - b + c)$
$\Rightarrow (a^2 + (-ab) + ac) + ((-ab) + b^2 + (-bc)) + (ac + (-bc) + c^2)$
$\Rightarrow a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2$
$\Rightarrow a^2 + b^2 + c^2 + (-ab - ab) + (-bc - bc) + (ac + ac)$
$\Rightarrow a^2 + b^2 + c^2 + (-2ab) + (-2bc) + (2ac)$
$\Rightarrow a^2 + b^2 + c^2 + (-2ab - 2bc + 2ac)$
$\Rightarrow a^2 + b^2 + c^2 - 2(ab - bc + ac)$ Ans
(vii) $(3x + \dfrac{1}{3x})^2$
$\Rightarrow (3x + \dfrac{1}{3x}) (3x + \dfrac{1}{3x})$
$\Rightarrow (3x)^2 + (2 \times 3x \times \dfrac{1}{3x}) + (\dfrac{1}{3x})^2$
$\Rightarrow 9x^2 + (\dfrac{6x}{3x}) + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + 2 + \dfrac{1}{9x^2}$ Ans
(viii) $(2x - \dfrac{1}{2x})^2$
$\Rightarrow (2x - \dfrac{1}{2x}) (2x - \dfrac{1}{2x})$
$\Rightarrow (2x)^2 + (2 \times 2x \times (-\dfrac{1}{2x})) + (\dfrac{1}{2x})^2$
$\Rightarrow 4x^2 + (-\dfrac{4x}{2x}) + \dfrac{1}{4x^2}$
$\Rightarrow 4x^2 - 2 + \dfrac{1}{4x^2}$ Ans
$\Rightarrow (2a + b) (2a + b)$
$\Rightarrow (2a)^2 + (2ab + 2ab) + (b)^2$
$\Rightarrow 4a^2 + 4ab + b^2$ Ans
(ii) $(a - 2b)^2$
$\Rightarrow (a - 2b) (a - 2b)$
$\Rightarrow (a)^2 + (-2ab - 2ab) + (2b)^2$
$\Rightarrow a^2 + (-4ab) + 2b^2$
$\Rightarrow a^2 - 4ab + 2b^2$ Ans
(iii) $(a + \dfrac{1}{2a})^2$
$\Rightarrow (a + \dfrac{1}{2a}) (a + \dfrac{1}{2a})$
$\Rightarrow (a)^2 + ((\dfrac{1}{2a})a + (\dfrac{1}{2a})a) + (\dfrac{1}{2a})^2$
$\Rightarrow a^2 + (\dfrac{1}{2} + \dfrac{1}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + (\dfrac{1 + 1}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + (\dfrac{2}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + 1 + \dfrac{1}{4a^2}$ Ans
The same problem can also be solved in the following manner:
$\Rightarrow (a + \dfrac{1}{2a}) (a + \dfrac{1}{2a})$
$\Rightarrow (a)^2 + (2 \times a \times (\dfrac{1}{2a})) + (\dfrac{1}{2a})^2$
$\Rightarrow a^2 + (\dfrac{2}{2}) + \dfrac{1}{4a^2}$
$\Rightarrow a^2 + 1 + \dfrac{1}{4a^2}$ Ans
(iv) $(2a - \dfrac{1}{a})^2$
$\Rightarrow (2a - \dfrac{1}{a}) (2a - \dfrac{1}{a})$
$\Rightarrow (2a)^2 + (2 \times 2a \times (-\dfrac{1}{a})) + (\dfrac{1}{a})^2$
$\Rightarrow 4a^2 + (-\dfrac{4a}{a}) + \dfrac{1}{a^2}$
$\Rightarrow 4a^2 - 4 + \dfrac{1}{a^2}$ Ans
(v) $(a + b - c)^2$
$\Rightarrow (a + b - c) (a + b - c)$
$\Rightarrow (a^2 + ab + (-ac)) + (ab + b^2 + (-bc)) + ((-ac) + (-bc) + c^2)$
$\Rightarrow a^2 + ab - ac + ab + b^2 - bc - ac - bc + c^2$
$\Rightarrow a^2 + b^2 + c^2 + (ab + ab) + (-bc - bc) + (-ac - ac)$
$\Rightarrow a^2 + b^2 + c^2 + (2ab) + (-2bc) + (-2ac)$
$\Rightarrow a^2 + b^2 + c^2 + (2ab - 2bc - 2ac)$
$\Rightarrow a^2 + b^2 + c^2 + 2(ab - bc - ac)$ Ans
(vi) $(a - b + c)^2$
$\Rightarrow (a - b + c) (a - b + c)$
$\Rightarrow (a^2 + (-ab) + ac) + ((-ab) + b^2 + (-bc)) + (ac + (-bc) + c^2)$
$\Rightarrow a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2$
$\Rightarrow a^2 + b^2 + c^2 + (-ab - ab) + (-bc - bc) + (ac + ac)$
$\Rightarrow a^2 + b^2 + c^2 + (-2ab) + (-2bc) + (2ac)$
$\Rightarrow a^2 + b^2 + c^2 + (-2ab - 2bc + 2ac)$
$\Rightarrow a^2 + b^2 + c^2 - 2(ab - bc + ac)$ Ans
(vii) $(3x + \dfrac{1}{3x})^2$
$\Rightarrow (3x + \dfrac{1}{3x}) (3x + \dfrac{1}{3x})$
$\Rightarrow (3x)^2 + (2 \times 3x \times \dfrac{1}{3x}) + (\dfrac{1}{3x})^2$
$\Rightarrow 9x^2 + (\dfrac{6x}{3x}) + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + 2 + \dfrac{1}{9x^2}$ Ans
(viii) $(2x - \dfrac{1}{2x})^2$
$\Rightarrow (2x - \dfrac{1}{2x}) (2x - \dfrac{1}{2x})$
$\Rightarrow (2x)^2 + (2 \times 2x \times (-\dfrac{1}{2x})) + (\dfrac{1}{2x})^2$
$\Rightarrow 4x^2 + (-\dfrac{4x}{2x}) + \dfrac{1}{4x^2}$
$\Rightarrow 4x^2 - 2 + \dfrac{1}{4x^2}$ Ans
Q7. Find the Square:
(i) $x + 3y$
(ii) $2x - 5y$
(iii) $a + \dfrac{1}{5a}$
(iv) $2a - \dfrac{1}{a}$
(v) $x - 2y + 1$
(vi) $3a - 2b - 5c$
(vii) $2x + \dfrac{1}{x} + 1$
(viii) $5 - x + \dfrac{2}{x}$
(ix) $2x - 3y + z$
(x) $x + \dfrac{1}{x} - 1$
Show Answer
(i) $x + 3y$
$\Rightarrow (x + 3y) (x + 3y)$
$\Rightarrow (x)^2 + (2 \times x \times 3y) + (3y)^2$
$\Rightarrow x^2 + 6x + 9y^2$ Ans
(ii) $2x - 5y$
$\Rightarrow (2x - 5y) (2x - 5y)$
$\Rightarrow (2x)^2 + (2 \times 2x \times (-5y)) + (5y)^2$
$\Rightarrow 4x^2 + 20xy + 25y^2$ Ans
(iii) $a + \dfrac{1}{5a}$
$\Rightarrow (a + \dfrac{1}{5a}) (a + \dfrac{1}{5a})$
$\Rightarrow (a)^2 + (2 \times a \times \dfrac{1}{5a}) + (\dfrac{1}{5a})^2$
$\Rightarrow a^2 + (\dfrac{2a}{5a}) + \dfrac{1}{25a^2}$
$\Rightarrow a^2 + \dfrac{2}{5} + \dfrac{1}{25a^2}$ Ans
(iv) $2a - \dfrac{1}{a}$
$\Rightarrow (2a - \dfrac{1}{a}) (2a - \dfrac{1}{a})$
$\Rightarrow (2a)^2 + (2 \times 2a \times (-\dfrac{1}{a})) + (\dfrac{1}{a})^2$
$\Rightarrow 4a^2 + (-\dfrac{4a}{a}) + \dfrac{1}{a^2}$
$\Rightarrow 4a^2 - 4 + \dfrac{1}{a^2}$ Ans
(v) $x - 2y + 1$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore x^2 + (2y)^2 + (1)^2 - 2(x \times 2y) - 2(2y \times 1) + 2(x \times 1)$
$\Rightarrow x^2 + 4y^2 + 1 - 2(2xy) - 2(2y) + 2(x)$
$\Rightarrow x^2 + 4y^2 + 1 - 4xy - 4y + 2x$ Ans
(vi) $3a - 2b - 5c$
$\because (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ac$
$\therefore (3a)^2 + (2b)^2 + (5c)^2 - 2(3a \times 2b) - 2(2b \times 5c) + 2(2b \times 5c)$
$\Rightarrow 9a^2 + 4b^2 + 25c^2 - 2(6ab) - 2(10bc) + 2(15ac)$
$\Rightarrow 9a^2 + 4b^2 + 25c^2 - 12ab + 20bc - 30ac$ Ans
(vii) $2x + \dfrac{1}{x} + 1$
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$
$\therefore (2x)^2 + (\dfrac{1}{x})^2 + (1)^2 + 2(2x \times \dfrac{1}{x}) + 2(\dfrac{1}{x} \times 1) + 2(2x \times 1)$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 1 + 2(2) + 2(\dfrac{1}{x}) + 2(2x)$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 1 + 4 + \dfrac{2}{x} + 4x$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 5 + \dfrac{2}{x} + 4x$ Ans
(viii) $5 - x + \dfrac{2}{x}$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore (5)^2 + (x)^2 + (\dfrac{2}{x})^2 - 2(5 \times x) - 2(x \times \dfrac{2}{x}) + 2(5 \times \dfrac{2}{x})$
$\Rightarrow 25 + x^2 + \dfrac{4}{x^2} - 2(5x) - 2(2) + 2(\dfrac{10}{x})$
$\Rightarrow 25 + x^2 + \dfrac{4}{x^2} - 10x - 4 + \dfrac{20}{x}$
$\Rightarrow (25 - 4) + x^2 + \dfrac{4}{x^2} - 10x + \dfrac{20}{x}$
$\Rightarrow 21 + x^2 + \dfrac{4}{x^2} - 10x + \dfrac{20}{x}$
(ix) $2x - 3y + z$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore (2x)^2 + (3y)^2 + (z)^2 - 2(2x \times 3y) - 2(3y \times z) + 2(2x \times z)$
$\Rightarrow 4x^2 + 9y^2 + z^2 - 2(6xy) - 2(3yz) + 2(2xz)$
$\Rightarrow 4x^2 + 9y^2 + z^2 - 12xy - 6yz + 4xz$ Ans
(x) $x + \dfrac{1}{x} - 1$
$\because (a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ac$
$\therefore (x)^2 + (\dfrac{1}{x})^2 + (1)^2 + 2(x \times \dfrac{1}{x}) - 2(\dfrac{1}{x} \times 1) - 2(x \times 1)$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 2(1) - 2(\dfrac{1}{x}) - 2(x)$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 2 - \dfrac{2}{x} - 2x$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 3 - \dfrac{2}{x} - 2x$ Ans
$\Rightarrow (x + 3y) (x + 3y)$
$\Rightarrow (x)^2 + (2 \times x \times 3y) + (3y)^2$
$\Rightarrow x^2 + 6x + 9y^2$ Ans
(ii) $2x - 5y$
$\Rightarrow (2x - 5y) (2x - 5y)$
$\Rightarrow (2x)^2 + (2 \times 2x \times (-5y)) + (5y)^2$
$\Rightarrow 4x^2 + 20xy + 25y^2$ Ans
(iii) $a + \dfrac{1}{5a}$
$\Rightarrow (a + \dfrac{1}{5a}) (a + \dfrac{1}{5a})$
$\Rightarrow (a)^2 + (2 \times a \times \dfrac{1}{5a}) + (\dfrac{1}{5a})^2$
$\Rightarrow a^2 + (\dfrac{2a}{5a}) + \dfrac{1}{25a^2}$
$\Rightarrow a^2 + \dfrac{2}{5} + \dfrac{1}{25a^2}$ Ans
(iv) $2a - \dfrac{1}{a}$
$\Rightarrow (2a - \dfrac{1}{a}) (2a - \dfrac{1}{a})$
$\Rightarrow (2a)^2 + (2 \times 2a \times (-\dfrac{1}{a})) + (\dfrac{1}{a})^2$
$\Rightarrow 4a^2 + (-\dfrac{4a}{a}) + \dfrac{1}{a^2}$
$\Rightarrow 4a^2 - 4 + \dfrac{1}{a^2}$ Ans
(v) $x - 2y + 1$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore x^2 + (2y)^2 + (1)^2 - 2(x \times 2y) - 2(2y \times 1) + 2(x \times 1)$
$\Rightarrow x^2 + 4y^2 + 1 - 2(2xy) - 2(2y) + 2(x)$
$\Rightarrow x^2 + 4y^2 + 1 - 4xy - 4y + 2x$ Ans
(vi) $3a - 2b - 5c$
$\because (a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ac$
$\therefore (3a)^2 + (2b)^2 + (5c)^2 - 2(3a \times 2b) - 2(2b \times 5c) + 2(2b \times 5c)$
$\Rightarrow 9a^2 + 4b^2 + 25c^2 - 2(6ab) - 2(10bc) + 2(15ac)$
$\Rightarrow 9a^2 + 4b^2 + 25c^2 - 12ab + 20bc - 30ac$ Ans
(vii) $2x + \dfrac{1}{x} + 1$
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$
$\therefore (2x)^2 + (\dfrac{1}{x})^2 + (1)^2 + 2(2x \times \dfrac{1}{x}) + 2(\dfrac{1}{x} \times 1) + 2(2x \times 1)$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 1 + 2(2) + 2(\dfrac{1}{x}) + 2(2x)$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 1 + 4 + \dfrac{2}{x} + 4x$
$\Rightarrow 4x^2 + \dfrac{1}{x^2} + 5 + \dfrac{2}{x} + 4x$ Ans
(viii) $5 - x + \dfrac{2}{x}$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore (5)^2 + (x)^2 + (\dfrac{2}{x})^2 - 2(5 \times x) - 2(x \times \dfrac{2}{x}) + 2(5 \times \dfrac{2}{x})$
$\Rightarrow 25 + x^2 + \dfrac{4}{x^2} - 2(5x) - 2(2) + 2(\dfrac{10}{x})$
$\Rightarrow 25 + x^2 + \dfrac{4}{x^2} - 10x - 4 + \dfrac{20}{x}$
$\Rightarrow (25 - 4) + x^2 + \dfrac{4}{x^2} - 10x + \dfrac{20}{x}$
$\Rightarrow 21 + x^2 + \dfrac{4}{x^2} - 10x + \dfrac{20}{x}$
(ix) $2x - 3y + z$
$\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ac$
$\therefore (2x)^2 + (3y)^2 + (z)^2 - 2(2x \times 3y) - 2(3y \times z) + 2(2x \times z)$
$\Rightarrow 4x^2 + 9y^2 + z^2 - 2(6xy) - 2(3yz) + 2(2xz)$
$\Rightarrow 4x^2 + 9y^2 + z^2 - 12xy - 6yz + 4xz$ Ans
(x) $x + \dfrac{1}{x} - 1$
$\because (a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ac$
$\therefore (x)^2 + (\dfrac{1}{x})^2 + (1)^2 + 2(x \times \dfrac{1}{x}) - 2(\dfrac{1}{x} \times 1) - 2(x \times 1)$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 2(1) - 2(\dfrac{1}{x}) - 2(x)$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 1 + 2 - \dfrac{2}{x} - 2x$
$\Rightarrow x^2 + \dfrac{1}{x^2} + 3 - \dfrac{2}{x} - 2x$ Ans
Q8. Evaluate using expansion of $(a + b)^2$ or $(a - b)^2$:
(i) $(208)^2$
(ii) $(92)^2$
(iii) $(415)^2$
(iv) $(188)^2$
(v) $(9.4)^2$
(vi) $(20.7)^2$
Show Answer
(i) $(208)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (208)^2 = (200 + 8)(200 + 8)$
$\Rightarrow (200)^2 + 2(200 \times 8) + (8)^2$
$\Rightarrow 40,000 + 2(1600) + 64$
$\Rightarrow 40,000 + 3200 + 64$
$\Rightarrow 43,200 + 64 = 43,264$ Ans
(ii) $(92)^2$
$\because (a - b)^2 = (a - b)(a - b)$
$\Rightarrow a^2 - 2ab + b^2$
$\therefore (92)^2 = (100 - 8)(100 - 8)$
$\Rightarrow (100)^2 - 2(100 \times 8) + (8)^2$
$\Rightarrow 10,000 - 2(800) + 64$
$\Rightarrow 10,000 - 1600 + 64$
$\Rightarrow 8,400 + 64 = 8,464$ Ans
(iii) $(415)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (415)^2 = (400 + 15)(400 + 15)$
$\Rightarrow (400)^2 + 2(400 \times 15) + (15)^2$
$\Rightarrow 1,60,000 + 2(6000) + 225$
$\Rightarrow 1,60,000 + 12000 + 225$
$\Rightarrow 1,72,000 + 225 = 1,72,225$ Ans
(iv) $(188)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (188)^2 = (180 + 8)(180 + 8)$
$\Rightarrow (180)^2 + 2(180 \times 8) + (8)^2$
$\Rightarrow 32,400 + 2(1440) + 64$
$\Rightarrow 32,400 + 2880 + 64$
$\Rightarrow 35,280 + 64 = 35,344$ Ans
(v) $(9.4)^2$
$\because (a - b)^2 = (a - b)(a - b)$
$\Rightarrow a^2 - 2ab + b^2$
$\therefore (9.4)^2 = (10 - 0.6)(10 - 0.6)$
$\Rightarrow (10)^2 - 2(10 \times 0.6) + (0.6)^2$
$\Rightarrow 100 - 2(6) + 0.36$
$\Rightarrow 100 - 12 + 0.36$
$\Rightarrow 88 + 0.36 = 88.36$ Ans
(vi) $(20.7)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (20.7)^2 = (20 + 0.7)(20 + 0.7)$
$\Rightarrow (20)^2 + 2(20 \times 0.7) + (0.7)^2$
$\Rightarrow 400 + 2(14) + 0.49$
$\Rightarrow 400 + 28 + 0.49$
$\Rightarrow 428 + 0.49 = 428.49$ Ans
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (208)^2 = (200 + 8)(200 + 8)$
$\Rightarrow (200)^2 + 2(200 \times 8) + (8)^2$
$\Rightarrow 40,000 + 2(1600) + 64$
$\Rightarrow 40,000 + 3200 + 64$
$\Rightarrow 43,200 + 64 = 43,264$ Ans
(ii) $(92)^2$
$\because (a - b)^2 = (a - b)(a - b)$
$\Rightarrow a^2 - 2ab + b^2$
$\therefore (92)^2 = (100 - 8)(100 - 8)$
$\Rightarrow (100)^2 - 2(100 \times 8) + (8)^2$
$\Rightarrow 10,000 - 2(800) + 64$
$\Rightarrow 10,000 - 1600 + 64$
$\Rightarrow 8,400 + 64 = 8,464$ Ans
(iii) $(415)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (415)^2 = (400 + 15)(400 + 15)$
$\Rightarrow (400)^2 + 2(400 \times 15) + (15)^2$
$\Rightarrow 1,60,000 + 2(6000) + 225$
$\Rightarrow 1,60,000 + 12000 + 225$
$\Rightarrow 1,72,000 + 225 = 1,72,225$ Ans
(iv) $(188)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (188)^2 = (180 + 8)(180 + 8)$
$\Rightarrow (180)^2 + 2(180 \times 8) + (8)^2$
$\Rightarrow 32,400 + 2(1440) + 64$
$\Rightarrow 32,400 + 2880 + 64$
$\Rightarrow 35,280 + 64 = 35,344$ Ans
(v) $(9.4)^2$
$\because (a - b)^2 = (a - b)(a - b)$
$\Rightarrow a^2 - 2ab + b^2$
$\therefore (9.4)^2 = (10 - 0.6)(10 - 0.6)$
$\Rightarrow (10)^2 - 2(10 \times 0.6) + (0.6)^2$
$\Rightarrow 100 - 2(6) + 0.36$
$\Rightarrow 100 - 12 + 0.36$
$\Rightarrow 88 + 0.36 = 88.36$ Ans
(vi) $(20.7)^2$
$\because (a + b)^2 = (a + b)(a + b)$
$\Rightarrow a^2 + 2ab + b^2$
$\therefore (20.7)^2 = (20 + 0.7)(20 + 0.7)$
$\Rightarrow (20)^2 + 2(20 \times 0.7) + (0.7)^2$
$\Rightarrow 400 + 2(14) + 0.49$
$\Rightarrow 400 + 28 + 0.49$
$\Rightarrow 428 + 0.49 = 428.49$ Ans
Q9. Expand:
(i) $(2a + b)^3$
(ii) $(a - 2b)^3$
(iii) $(3x - 2y)^3$
(iv) $(x + 5y)^3$
(v) $(a + \dfrac{1}{a})^3$
(vi) $(2a - \dfrac{1}{2a})^3$
Show Answer
(i) $(2a + b)^3$
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$ used in expAnsions
$\Rightarrow a^3 + b^3 + 3ab(a + b)$ used in formulas
$\therefore (2a + b)^3 = (2a)^3 + 3((2a)^2b) + 3((2a)b^2) + b^3$
$\Rightarrow 8a^3 + 3(4a^2b) + 3((2ab^2) + b^3$
$\Rightarrow 8a^3 + 12a^2b + 6ab^2 + b^3$ Ans
(ii) $(a - 2b)^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$ used in expAnsions
$\Rightarrow a^3 - b^3 - 3ab(a - b)$ used in formulas
$\therefore (a - 2b)^3 = a^3 - 3(a^2(2b)) + 3(a(2b))^2 - (2b)^3$
$\Rightarrow a^3 - 3(a^2(2b)) + 3(a(4b^2)) - (2b)^3$
$\Rightarrow a^3 - 3(2a^2b) + 3(4ab^2) - 8b^3$
$\Rightarrow a^3 - 6a^2b + 12ab^2 - 8b^3$ Ans
(iii) $(3x - 2y)^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (3x - 2y)^3 = (3x)^3 - 3((3x)^2(2y)) + 3((3x)(2y))^2 - (2y)^3$
$\Rightarrow 27x^3 - 3(9x^2(2y)) + 3((3x)4y^2) - 8y^3$
$\Rightarrow 27x^3 - 3(18x^2y) + 3(12xy^2) - 8y^3$
$\Rightarrow 27x^3 - 54x^2y + 36xy^2 - 8y^3$ Ans
(iv) $(x + 5y)^3$
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (x + 5y)^3 = (x)^3 + 3((x)^2(5y)) + 3((x)(5y)^2) + (5y)^3$
$\Rightarrow x^3 + 3(5x^2y) + 3((x)(25y^2) + 125y^3$
$\Rightarrow x^3 + 15x^2y + 3((25xy^2) + 125y^3$
$\Rightarrow x^3 + 15x^2y + 75xy^2 + 125y^3$ Ans
(v) $(a + \dfrac{1}{a})^3$
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (a + \dfrac{1}{a})^3 = (a)^3 + 3((a)^2(\dfrac{1}{a})) + 3((a)(\dfrac{1}{a})^2) + (\dfrac{1}{a})^3$
$\Rightarrow a^3 + 3(\dfrac{a^2}{a}) + 3((a)(\dfrac{1}{a^2})) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3(a) + 3(\dfrac{a}{a^2}) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3a + 3(\dfrac{1}{a}) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3a + \dfrac{3}{a} + \dfrac{1}{a^3}$ Ans
(vi) $(2a - \dfrac{1}{2a})^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (2a - \dfrac{1}{2a})^3 = (2a)^3 - 3((2a)^2(\dfrac{1}{2a})) + 3((2a)(\dfrac{1}{2a}))^2 - (\dfrac{1}{2a})^3$
$\Rightarrow 8a^3 - 3((4a^2)(\dfrac{1}{2a})) + 3((2a)(\dfrac{1}{4a^2})) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 3(\dfrac{4a^2}{2a}) + 3(\dfrac{2a}{4a^2}) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 3(2a) + 3(\dfrac{1}{2a}) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 6a + \dfrac{3}{2a} - \dfrac{1}{8a^3}$ Ans
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$ used in expAnsions
$\Rightarrow a^3 + b^3 + 3ab(a + b)$ used in formulas
$\therefore (2a + b)^3 = (2a)^3 + 3((2a)^2b) + 3((2a)b^2) + b^3$
$\Rightarrow 8a^3 + 3(4a^2b) + 3((2ab^2) + b^3$
$\Rightarrow 8a^3 + 12a^2b + 6ab^2 + b^3$ Ans
(ii) $(a - 2b)^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$ used in expAnsions
$\Rightarrow a^3 - b^3 - 3ab(a - b)$ used in formulas
$\therefore (a - 2b)^3 = a^3 - 3(a^2(2b)) + 3(a(2b))^2 - (2b)^3$
$\Rightarrow a^3 - 3(a^2(2b)) + 3(a(4b^2)) - (2b)^3$
$\Rightarrow a^3 - 3(2a^2b) + 3(4ab^2) - 8b^3$
$\Rightarrow a^3 - 6a^2b + 12ab^2 - 8b^3$ Ans
(iii) $(3x - 2y)^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (3x - 2y)^3 = (3x)^3 - 3((3x)^2(2y)) + 3((3x)(2y))^2 - (2y)^3$
$\Rightarrow 27x^3 - 3(9x^2(2y)) + 3((3x)4y^2) - 8y^3$
$\Rightarrow 27x^3 - 3(18x^2y) + 3(12xy^2) - 8y^3$
$\Rightarrow 27x^3 - 54x^2y + 36xy^2 - 8y^3$ Ans
(iv) $(x + 5y)^3$
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (x + 5y)^3 = (x)^3 + 3((x)^2(5y)) + 3((x)(5y)^2) + (5y)^3$
$\Rightarrow x^3 + 3(5x^2y) + 3((x)(25y^2) + 125y^3$
$\Rightarrow x^3 + 15x^2y + 3((25xy^2) + 125y^3$
$\Rightarrow x^3 + 15x^2y + 75xy^2 + 125y^3$ Ans
(v) $(a + \dfrac{1}{a})^3$
$\because (a + b)^3 = (a + b) (a + b)^2$
$\Rightarrow (a + b) (a^2 + 2ab + b^2)$
$\Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)$
$\Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$
$\Rightarrow a^3 + (2a^2b + a^2b) + (2ab^2 + ab^2 ) + b^3$
$\Rightarrow a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (a + \dfrac{1}{a})^3 = (a)^3 + 3((a)^2(\dfrac{1}{a})) + 3((a)(\dfrac{1}{a})^2) + (\dfrac{1}{a})^3$
$\Rightarrow a^3 + 3(\dfrac{a^2}{a}) + 3((a)(\dfrac{1}{a^2})) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3(a) + 3(\dfrac{a}{a^2}) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3a + 3(\dfrac{1}{a}) + \dfrac{1}{a^3}$
$\Rightarrow a^3 + 3a + \dfrac{3}{a} + \dfrac{1}{a^3}$ Ans
(vi) $(2a - \dfrac{1}{2a})^3$
$\because (a - b)^3 = (a - b) (a - b)^2$
$\Rightarrow (a - b) (a^2 - 2ab + b^2)$
$\Rightarrow a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2)$
$\Rightarrow a^3 - 2a^2b + ab^2 - a^2b + 2ab^2 - b^3$
$\Rightarrow a^3 + (-2a^2b - a^2b) + (2ab^2 + ab^2 ) - b^3$
$\Rightarrow a^3 + (-3a^2b) + 3ab^2 - b^3$
$\Rightarrow a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (2a - \dfrac{1}{2a})^3 = (2a)^3 - 3((2a)^2(\dfrac{1}{2a})) + 3((2a)(\dfrac{1}{2a}))^2 - (\dfrac{1}{2a})^3$
$\Rightarrow 8a^3 - 3((4a^2)(\dfrac{1}{2a})) + 3((2a)(\dfrac{1}{4a^2})) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 3(\dfrac{4a^2}{2a}) + 3(\dfrac{2a}{4a^2}) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 3(2a) + 3(\dfrac{1}{2a}) - \dfrac{1}{8a^3}$
$\Rightarrow 8a^3 - 6a + \dfrac{3}{2a} - \dfrac{1}{8a^3}$ Ans
Q10. Find the cube of:
(i) $a + 2$
(ii) $2a - 1$
(iii) $2a + 3b$
(iv) $3b - 2a$
(v) $2x + \dfrac{1}{x}$
(vi) $x - \dfrac{1}{2}$
Show Answer
(i) $a + 2$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (a + 2)^3 = (a)^3 + 3((a)^2(2)) + 3((a)(2)^2) + (2)^3$
$\Rightarrow a^3 + 3((a^2)(2)) + 3((a)(4)) + 8$
$\Rightarrow a^3 + 3(2a^2) + 3(4a) + 8$
$\Rightarrow a^3 + 6a^2 + 12a + 8$ Ans
(ii) $2a - 1$
$\because (a - 1)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (a - 1)^3 = (a)^3 - 3((a)^2(1)) + 3((a)(1)^2) - (1)^3$
$\Rightarrow a^3 - 3((a^2)(1)) + 3((a)(1)) - 1$
$\Rightarrow a^3 - 3(a^2) + 3(a) - 1$
$\Rightarrow a^3 - 3a^2 + 3a - 1$ Ans
(iii) $2a + 3b$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (2a + 3b)^3 = (2a)^3 + 3((2a)^2(3b)) + 3((2a)(3b)^2) + (3b)^3$
$\Rightarrow 8a^3 + 3((4a^2)(3b)) + 3((2a)(9b^2)) + 27b^3$
$\Rightarrow 8a^3 + 3(12a^2b) + 3(18ab^2) + 27b^3$
$\Rightarrow 8a^3 + 36a^2b + 54ab^2 + 27b^3$ Ans
(iv) $3b - 2a$
$\because (a - 1)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (3b - 2a)^3 = (3b)^3 - 3((3b)^2(2a)) + 3((3b)(2a)^2) - (2a)^3$
$\Rightarrow 27b^3 - 3((9b^2)(2a)) + 3((3b)(4a^2)) - 8a^3$
$\Rightarrow 27b^3 - 3(18b^2a) + 3(12ba^2) - 8a^3$
$\Rightarrow 27b^3 - 54b^2a + 36ba^2 - 8a^3$ Ans
(v) $2x + \dfrac{1}{x}$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (2x + \dfrac{1}{x})^3 = (2x)^3 + 3((2x)^2(\dfrac{1}{x})) + 3((2x)(\dfrac{1}{x})^2) + (\dfrac{1}{x})^3$
$\Rightarrow 8x^3 + 3((4x^2)(\dfrac{1}{x})) + 3((2x)(\dfrac{1}{x^2})) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 3(\dfrac{4x^2}{x}) + 3(\dfrac{2x}{x^2}) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 3(4x) + 3(\dfrac{2}{x}) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 12x + \dfrac{6}{x} + \dfrac{1}{x^3}$ Ans
(vi) $x - \dfrac{1}{2}$
$\because (x - \dfrac{1}{2})^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (a - 1)^3 = (x)^3 - 3((x)^2(\dfrac{1}{2})) + 3((x)(\dfrac{1}{2})^2) - (\dfrac{1}{2})^3$
$\Rightarrow x^3 - 3((x^2)(\dfrac{1}{2})) + 3((x)(\dfrac{1}{4})) - \dfrac{1}{8}$
$\Rightarrow x^3 - 3(\dfrac{x^2}{2}) + 3(\dfrac{x}{4}) - \dfrac{1}{8}$
$\Rightarrow x^3 - \dfrac{3x^2}{2} + \dfrac{3x}{4} - \dfrac{1}{8}$ Ans
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (a + 2)^3 = (a)^3 + 3((a)^2(2)) + 3((a)(2)^2) + (2)^3$
$\Rightarrow a^3 + 3((a^2)(2)) + 3((a)(4)) + 8$
$\Rightarrow a^3 + 3(2a^2) + 3(4a) + 8$
$\Rightarrow a^3 + 6a^2 + 12a + 8$ Ans
(ii) $2a - 1$
$\because (a - 1)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (a - 1)^3 = (a)^3 - 3((a)^2(1)) + 3((a)(1)^2) - (1)^3$
$\Rightarrow a^3 - 3((a^2)(1)) + 3((a)(1)) - 1$
$\Rightarrow a^3 - 3(a^2) + 3(a) - 1$
$\Rightarrow a^3 - 3a^2 + 3a - 1$ Ans
(iii) $2a + 3b$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (2a + 3b)^3 = (2a)^3 + 3((2a)^2(3b)) + 3((2a)(3b)^2) + (3b)^3$
$\Rightarrow 8a^3 + 3((4a^2)(3b)) + 3((2a)(9b^2)) + 27b^3$
$\Rightarrow 8a^3 + 3(12a^2b) + 3(18ab^2) + 27b^3$
$\Rightarrow 8a^3 + 36a^2b + 54ab^2 + 27b^3$ Ans
(iv) $3b - 2a$
$\because (a - 1)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (3b - 2a)^3 = (3b)^3 - 3((3b)^2(2a)) + 3((3b)(2a)^2) - (2a)^3$
$\Rightarrow 27b^3 - 3((9b^2)(2a)) + 3((3b)(4a^2)) - 8a^3$
$\Rightarrow 27b^3 - 3(18b^2a) + 3(12ba^2) - 8a^3$
$\Rightarrow 27b^3 - 54b^2a + 36ba^2 - 8a^3$ Ans
(v) $2x + \dfrac{1}{x}$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (2x + \dfrac{1}{x})^3 = (2x)^3 + 3((2x)^2(\dfrac{1}{x})) + 3((2x)(\dfrac{1}{x})^2) + (\dfrac{1}{x})^3$
$\Rightarrow 8x^3 + 3((4x^2)(\dfrac{1}{x})) + 3((2x)(\dfrac{1}{x^2})) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 3(\dfrac{4x^2}{x}) + 3(\dfrac{2x}{x^2}) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 3(4x) + 3(\dfrac{2}{x}) + \dfrac{1}{x^3}$
$\Rightarrow 8x^3 + 12x + \dfrac{6}{x} + \dfrac{1}{x^3}$ Ans
(vi) $x - \dfrac{1}{2}$
$\because (x - \dfrac{1}{2})^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (a - 1)^3 = (x)^3 - 3((x)^2(\dfrac{1}{2})) + 3((x)(\dfrac{1}{2})^2) - (\dfrac{1}{2})^3$
$\Rightarrow x^3 - 3((x^2)(\dfrac{1}{2})) + 3((x)(\dfrac{1}{4})) - \dfrac{1}{8}$
$\Rightarrow x^3 - 3(\dfrac{x^2}{2}) + 3(\dfrac{x}{4}) - \dfrac{1}{8}$
$\Rightarrow x^3 - \dfrac{3x^2}{2} + \dfrac{3x}{4} - \dfrac{1}{8}$ Ans
Q11. If $a + b = 5$ and $ab = 6$, find $a^2 + b^2$
Show Answer
$\because (a + b)^2 = a^2 + b^2 + 2ab$
When using in formulas, $a^2 + 2ab + b^2$ is used as $a^2 + b^2 + 2ab$
Given $a + b = 5$ and $ab = 6$, we have:
$\therefore (5)^2 = a^2 + b^2 + 2 \times 6$
$\Rightarrow 25 = a^2 + b^2 + 12$
$\Rightarrow 25 - 12 = a^2 + b^2$
$\Rightarrow a^2 + b^2 = 13$ Ans
When using in formulas, $a^2 + 2ab + b^2$ is used as $a^2 + b^2 + 2ab$
Given $a + b = 5$ and $ab = 6$, we have:
$\therefore (5)^2 = a^2 + b^2 + 2 \times 6$
$\Rightarrow 25 = a^2 + b^2 + 12$
$\Rightarrow 25 - 12 = a^2 + b^2$
$\Rightarrow a^2 + b^2 = 13$ Ans
Q12. If $a - b = 6$ and $ab = 16$, find $a^2 + b^2$
Show Answer
$\because (a - b)^2 = a^2 + b^2 - 2ab$
When using in formulas, $a^2 - 2ab + b^2$ is used as $a^2 + b^2 - 2ab$
Given $a - b = 6$ and $ab = 16$, we have:
$\therefore (6)^2 = a^2 + b^2 - 2 \times 16$
$\Rightarrow 36 = a^2 + b^2 - 32$
$\Rightarrow 36 + 32 = a^2 + b^2$
$\Rightarrow a^2 + b^2 = 68$ Ans
When using in formulas, $a^2 - 2ab + b^2$ is used as $a^2 + b^2 - 2ab$
Given $a - b = 6$ and $ab = 16$, we have:
$\therefore (6)^2 = a^2 + b^2 - 2 \times 16$
$\Rightarrow 36 = a^2 + b^2 - 32$
$\Rightarrow 36 + 32 = a^2 + b^2$
$\Rightarrow a^2 + b^2 = 68$ Ans
Q13. If $a^2 + b^2 = 29$ and $ab = 10$, find:
(i) $a + b$
(ii) $a - b$
Show Answer
(i) $a + b$
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 29$ and $ab = 10$, we have:
$\therefore (a + b)^2 = 29 + 2 \times 10$
$\Rightarrow (a + b)^2 = 29 + 20$
$\Rightarrow (a + b)^2 = 49$
$\Rightarrow a + b = \pm\sqrt{49} = \pm7$ Ans
Note, the sign $\pm$ (plus-minus) suggests the possiblity of a Plus or Minus. In the above answer it means either $+7$ or $-7$.
(ii) $a - b$
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 29$ and $ab = 10$, we have:
$\therefore (a - b)^2 = 29 - 2 \times 10$
$\Rightarrow (a + b)^2 = 29 - 20$
$\Rightarrow (a + b)^2 = 9$
$\Rightarrow a + b = \pm\sqrt{9} = \pm3$ Ans
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 29$ and $ab = 10$, we have:
$\therefore (a + b)^2 = 29 + 2 \times 10$
$\Rightarrow (a + b)^2 = 29 + 20$
$\Rightarrow (a + b)^2 = 49$
$\Rightarrow a + b = \pm\sqrt{49} = \pm7$ Ans
Note, the sign $\pm$ (plus-minus) suggests the possiblity of a Plus or Minus. In the above answer it means either $+7$ or $-7$.
(ii) $a - b$
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 29$ and $ab = 10$, we have:
$\therefore (a - b)^2 = 29 - 2 \times 10$
$\Rightarrow (a + b)^2 = 29 - 20$
$\Rightarrow (a + b)^2 = 9$
$\Rightarrow a + b = \pm\sqrt{9} = \pm3$ Ans
Q14. If $a^2 + b^2 = 10$ and $ab = 3$; find:
(i) $a - b$
(ii) $a + b$
Show Answer
(i) $a - b$
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 10$ and $ab = 3$, we have:
$\therefore (a - b)^2 = 10 - 2 \times 3$
$\Rightarrow (a + b)^2 = 10 - 6$
$\Rightarrow (a + b)^2 = 4$
$\Rightarrow a + b = \pm\sqrt{4} = \pm2$ Ans
(ii) $a + b$
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 10$ and $ab = 3$, we have:
$\therefore (a + b)^2 = 10 + 2 \times 3$
$\Rightarrow (a + b)^2 = 10 + 6$
$\Rightarrow (a + b)^2 = 16$
$\Rightarrow a + b = \pm\sqrt{16} = \pm4$ Ans
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 10$ and $ab = 3$, we have:
$\therefore (a - b)^2 = 10 - 2 \times 3$
$\Rightarrow (a + b)^2 = 10 - 6$
$\Rightarrow (a + b)^2 = 4$
$\Rightarrow a + b = \pm\sqrt{4} = \pm2$ Ans
(ii) $a + b$
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 10$ and $ab = 3$, we have:
$\therefore (a + b)^2 = 10 + 2 \times 3$
$\Rightarrow (a + b)^2 = 10 + 6$
$\Rightarrow (a + b)^2 = 16$
$\Rightarrow a + b = \pm\sqrt{16} = \pm4$ Ans
Q15. If $a + \dfrac{1}{a} = 3$, find: $a^2 + \dfrac{1}{a^2}$
Show Answer
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a + \dfrac{1}{a} = 3$, we have:
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 7$ Ans
Given $a + \dfrac{1}{a} = 3$, we have:
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 9 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 7$ Ans