Q31. Find square of:
(i) $3x + \dfrac{2}{y}$
(ii) $\dfrac{5a}{6b} - \dfrac{6b}{5a}$
(iii) $2m^2 - \dfrac{2}{3}n^2$
(iv) $5x + \dfrac{1}{5x}$
(v) $8x + \dfrac{3}{2}y$
Show Answer
(i) $(3x + \dfrac{2}{y})^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (3x)^2 + 2(3x \times \dfrac{2}{y}) + (\dfrac{2}{y})^2$
$\Rightarrow 9x^2 + 2(\dfrac{6x}{y}) + \dfrac{4}{y^2}$
$\Rightarrow 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}$ Ans
(ii) $(\dfrac{5a}{6b} - \dfrac{6b}{5a})^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (\dfrac{5a}{6b})^2 - 2(\dfrac{5a}{6b} \times \dfrac{6b}{5a}) + (\dfrac{6b}{5a})^2$
$\Rightarrow \dfrac{25a^2}{36b^2} - 2(1) + \dfrac{36b^2}{25a^2}$
$\Rightarrow \dfrac{25a^2}{36b^2} - 2 + \dfrac{36b^2}{25a^2}$ Ans
(iii) $2m^2 - \dfrac{2}{3}n^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (2m^2)^2 - 2(2m^2 \times \dfrac{2}{3}n^2) + (\dfrac{2}{3}n^2)^2$
$\Rightarrow 4m^4 - 2(\dfrac{4m^2n^2}{3}) + \dfrac{4}{9}n^4$
$\Rightarrow 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4$ Ans
(iv) $5x + \dfrac{1}{5x}$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (5x)^2 + 2(5x \times \dfrac{1}{5x}) + (\dfrac{1}{5x})^2$
$\Rightarrow 25x^2 + 2(1) + \dfrac{1}{25x^2}$
$\Rightarrow 25x^2 + 2 + \dfrac{1}{25x^2}$ Ans
(v) $8x + \dfrac{3}{2}y$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (8x)^2 + 2(8x \times \dfrac{3}{2}y) + (\dfrac{3}{2}y)^2$
$\Rightarrow 64x^2 + 2(4x \times 3y) + \dfrac{9}{4}y^2$
$\Rightarrow 64x^2 + 2(12xy) + \dfrac{9}{4}y^2$
$\Rightarrow 64x^2 + 24xy + \dfrac{9}{4}y^2$ Ans
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (3x)^2 + 2(3x \times \dfrac{2}{y}) + (\dfrac{2}{y})^2$
$\Rightarrow 9x^2 + 2(\dfrac{6x}{y}) + \dfrac{4}{y^2}$
$\Rightarrow 9x^2 + \dfrac{12x}{y} + \dfrac{4}{y^2}$ Ans
(ii) $(\dfrac{5a}{6b} - \dfrac{6b}{5a})^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (\dfrac{5a}{6b})^2 - 2(\dfrac{5a}{6b} \times \dfrac{6b}{5a}) + (\dfrac{6b}{5a})^2$
$\Rightarrow \dfrac{25a^2}{36b^2} - 2(1) + \dfrac{36b^2}{25a^2}$
$\Rightarrow \dfrac{25a^2}{36b^2} - 2 + \dfrac{36b^2}{25a^2}$ Ans
(iii) $2m^2 - \dfrac{2}{3}n^2$
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (2m^2)^2 - 2(2m^2 \times \dfrac{2}{3}n^2) + (\dfrac{2}{3}n^2)^2$
$\Rightarrow 4m^4 - 2(\dfrac{4m^2n^2}{3}) + \dfrac{4}{9}n^4$
$\Rightarrow 4m^4 - \dfrac{8}{3}m^2n^2 + \dfrac{4}{9}n^4$ Ans
(iv) $5x + \dfrac{1}{5x}$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (5x)^2 + 2(5x \times \dfrac{1}{5x}) + (\dfrac{1}{5x})^2$
$\Rightarrow 25x^2 + 2(1) + \dfrac{1}{25x^2}$
$\Rightarrow 25x^2 + 2 + \dfrac{1}{25x^2}$ Ans
(v) $8x + \dfrac{3}{2}y$
$\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (8x)^2 + 2(8x \times \dfrac{3}{2}y) + (\dfrac{3}{2}y)^2$
$\Rightarrow 64x^2 + 2(4x \times 3y) + \dfrac{9}{4}y^2$
$\Rightarrow 64x^2 + 2(12xy) + \dfrac{9}{4}y^2$
$\Rightarrow 64x^2 + 24xy + \dfrac{9}{4}y^2$ Ans
Q32. If $a + \dfrac{1}{a} = 2$, find:
(i) $a^2 + \dfrac{1}{a^2}$
(ii) $a^4 + \dfrac{1}{a^4}$
Show Answer
(i) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a + \dfrac{1}{a} = 2$:
$\therefore (2)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 4 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 4 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 2$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From previous (i) we have $a^2 + \dfrac{1}{a^2} = 2$:
$\therefore (2)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 4 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 4 - 2 = a^4 + \dfrac{1}{a^4}$
$\Rightarrow a^4 + \dfrac{1}{a^4} = 2$ Ans
Given $a + \dfrac{1}{a} = 2$:
$\therefore (2)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 4 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 4 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 2$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From previous (i) we have $a^2 + \dfrac{1}{a^2} = 2$:
$\therefore (2)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 4 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 4 - 2 = a^4 + \dfrac{1}{a^4}$
$\Rightarrow a^4 + \dfrac{1}{a^4} = 2$ Ans
Q33. If $m - \dfrac{1}{m} = 5$, find:
(i) $m^2 + \dfrac{1}{m^2}$
(ii) $m^4 + \dfrac{1}{m^4}$
(iii) $m^2 - \dfrac{1}{m^2}$
Show Answer
(i) $\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
$\therefore (m - \dfrac{1}{m})^2 = m^2 + \dfrac{1}{m^2} - 2$
Given $m - \dfrac{1}{m} = 5$:
$\therefore (5)^2 = m^2 + \dfrac{1}{m^2} - 2$
$\Rightarrow 25 = m^2 + \dfrac{1}{m^2} - 2$
$\Rightarrow 25 + 2 = m^2 + \dfrac{1}{m^2}$
$\Rightarrow m^2 + \dfrac{1}{m^2} = 27$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
$\therefore (m^2 + \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} + 2$
From (i) we have $m^2 + \dfrac{1}{m^2} = 27$:
$\therefore (27)^2 = m^4 + \dfrac{1}{m^4} + 2$
$\Rightarrow 729 = m^4 + \dfrac{1}{m^4} + 2$
$\Rightarrow 729 - 2 = m^2 + \dfrac{1}{m^2}$
$\Rightarrow m^4 + \dfrac{1}{m^4} = 727$ Ans
(iii) $\because (a^2 - \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} - 2$
$\therefore (m^2 - \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} - 2$
From (ii) we have $m^4 - \dfrac{1}{m^4} = 727$:
$\therefore (m^2 - \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} - 2$
$\Rightarrow (m^2 - \dfrac{1}{m^2})^2 = 727 - 2$
$\Rightarrow (m^2 - \dfrac{1}{m^2})^2 = 725$
$\Rightarrow m^2 - \dfrac{1}{m^2} = \sqrt{725}$
$\Rightarrow m^2 - \dfrac{1}{m^2} = 26.925824036$ or $\sqrt[5]{29}$ Ans
$\therefore (m - \dfrac{1}{m})^2 = m^2 + \dfrac{1}{m^2} - 2$
Given $m - \dfrac{1}{m} = 5$:
$\therefore (5)^2 = m^2 + \dfrac{1}{m^2} - 2$
$\Rightarrow 25 = m^2 + \dfrac{1}{m^2} - 2$
$\Rightarrow 25 + 2 = m^2 + \dfrac{1}{m^2}$
$\Rightarrow m^2 + \dfrac{1}{m^2} = 27$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
$\therefore (m^2 + \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} + 2$
From (i) we have $m^2 + \dfrac{1}{m^2} = 27$:
$\therefore (27)^2 = m^4 + \dfrac{1}{m^4} + 2$
$\Rightarrow 729 = m^4 + \dfrac{1}{m^4} + 2$
$\Rightarrow 729 - 2 = m^2 + \dfrac{1}{m^2}$
$\Rightarrow m^4 + \dfrac{1}{m^4} = 727$ Ans
(iii) $\because (a^2 - \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} - 2$
$\therefore (m^2 - \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} - 2$
From (ii) we have $m^4 - \dfrac{1}{m^4} = 727$:
$\therefore (m^2 - \dfrac{1}{m^2})^2 = m^4 + \dfrac{1}{m^4} - 2$
$\Rightarrow (m^2 - \dfrac{1}{m^2})^2 = 727 - 2$
$\Rightarrow (m^2 - \dfrac{1}{m^2})^2 = 725$
$\Rightarrow m^2 - \dfrac{1}{m^2} = \sqrt{725}$
$\Rightarrow m^2 - \dfrac{1}{m^2} = 26.925824036$ or $\sqrt[5]{29}$ Ans
Q34. If $a^2 + b^2 = 41$ and $ab = 4$, find:
(i) $a - b$
(ii) $a + b$
Show Answer
(i) $a - b$
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 41$ and $ab = 4$:
$\therefore (a - b)^2 = 41 - 2(4)$
$\Rightarrow (a - b)^2 = 41 - 8$
$\Rightarrow (a - b)^2 = 33$
$\Rightarrow a - b = \sqrt{33}$ Ans
(ii) $a + b$
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 41$ and $ab = 4$:
$\therefore (a + b)^2 = 41 + 2(4)$
$\Rightarrow (a + b)^2 = 41 + 8$
$\Rightarrow (a + b)^2 = 49$
$\Rightarrow a + b = \sqrt{49} = 7$ Ans
$\because (a - b)^2 = a^2 + b^2 - 2ab$
Given $a^2 + b^2 = 41$ and $ab = 4$:
$\therefore (a - b)^2 = 41 - 2(4)$
$\Rightarrow (a - b)^2 = 41 - 8$
$\Rightarrow (a - b)^2 = 33$
$\Rightarrow a - b = \sqrt{33}$ Ans
(ii) $a + b$
$\because (a + b)^2 = a^2 + b^2 + 2ab$
Given $a^2 + b^2 = 41$ and $ab = 4$:
$\therefore (a + b)^2 = 41 + 2(4)$
$\Rightarrow (a + b)^2 = 41 + 8$
$\Rightarrow (a + b)^2 = 49$
$\Rightarrow a + b = \sqrt{49} = 7$ Ans
Q35. If $2a + \dfrac{1}{2a} = 8$, find:
(i) $4a^2 + \dfrac{1}{4a^2}$
(ii) $16a^4 + \dfrac{1}{16a^4}$
Show Answer
(i) $4a^2 + \dfrac{1}{4a^2}$
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $2a + \dfrac{1}{2a} = 8$:
$\therefore (2a + \dfrac{1}{2a})^2 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow (8)^2 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow 64 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow 64 - 2 = 4a^2 + \dfrac{1}{4a^2}$
$\Rightarrow 4a^2 + \dfrac{1}{4a^2} = 62$ Ans
(ii) $16a^4 + \dfrac{1}{16a^4}$
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
From (i) we have $4a^2 + \dfrac{1}{4a^2} = 62$:
$\therefore (4a^2 + \dfrac{1}{4a^2})^2 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow (62)^2 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow 3844 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow 3844 - 2 = 16a^2 + \dfrac{1}{16a^2}$
$\Rightarrow 16a^2 + \dfrac{1}{16a^2} = 3842$ Ans
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $2a + \dfrac{1}{2a} = 8$:
$\therefore (2a + \dfrac{1}{2a})^2 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow (8)^2 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow 64 = 4a^2 + \dfrac{1}{4a^2} + 2$
$\Rightarrow 64 - 2 = 4a^2 + \dfrac{1}{4a^2}$
$\Rightarrow 4a^2 + \dfrac{1}{4a^2} = 62$ Ans
(ii) $16a^4 + \dfrac{1}{16a^4}$
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
From (i) we have $4a^2 + \dfrac{1}{4a^2} = 62$:
$\therefore (4a^2 + \dfrac{1}{4a^2})^2 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow (62)^2 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow 3844 = 16a^2 + \dfrac{1}{16a^2} + 2$
$\Rightarrow 3844 - 2 = 16a^2 + \dfrac{1}{16a^2}$
$\Rightarrow 16a^2 + \dfrac{1}{16a^2} = 3842$ Ans
Q36. If $3x - \dfrac{1}{3x} = 5$, find:
(i) $9x^2 + \dfrac{1}{9x^2}$
(ii) $81x^4 + \dfrac{1}{81x^4}$
Show Answer
(i) $9x^2 + \dfrac{1}{9x^2}$
$\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $3x + \dfrac{1}{3x} = 5$:
$\therefore (3x - \dfrac{1}{3x})^2 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow (5)^2 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow 25 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow 25 + 2 = 9x^2 + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + \dfrac{1}{9x^2} = 27$ Ans
(ii) $81x^4 + \dfrac{1}{81x^4}$
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
From (i) we have $9x^2 + \dfrac{1}{9x^2} = 27$:
$\therefore (9x^2 + \dfrac{1}{9x^2})^2 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow (27)^2 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow 729 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow 729 - 2 = 81x^4 + \dfrac{1}{81x^4}$
$\Rightarrow 81x^4 + \dfrac{1}{81x^4} = 727$ Ans
$\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $3x + \dfrac{1}{3x} = 5$:
$\therefore (3x - \dfrac{1}{3x})^2 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow (5)^2 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow 25 = 9x^2 + \dfrac{1}{9x^2} - 2$
$\Rightarrow 25 + 2 = 9x^2 + \dfrac{1}{9x^2}$
$\Rightarrow 9x^2 + \dfrac{1}{9x^2} = 27$ Ans
(ii) $81x^4 + \dfrac{1}{81x^4}$
$\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
From (i) we have $9x^2 + \dfrac{1}{9x^2} = 27$:
$\therefore (9x^2 + \dfrac{1}{9x^2})^2 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow (27)^2 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow 729 = 81x^4 + \dfrac{1}{81x^4} + 2$
$\Rightarrow 729 - 2 = 81x^4 + \dfrac{1}{81x^4}$
$\Rightarrow 81x^4 + \dfrac{1}{81x^4} = 727$ Ans
Q37. Expand:
(i) $(3x - 4y + 5z)^2$
(ii) $(2a - 5b - 4c)^2$
(iii) $(5x + 3y)^3$
(iv) $(6a - 7b)^3$
Show Answer
(i) $(3x - 4y + 5z)^2$
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (3x)^2 + (-4y)^2 + (5z)^2 + 2((3x \times (-4y)) + ((-4y) \times 5z) + (5z \times 3x))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 + 2((-12xy) + (-20yz) + (15xz))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 + 2(-12xy) + 2(-20yz) + 2(15xz))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 - 24xy - 40yz + 30xz$ Ans
(ii) $(2a - 5b - 4c)^2$
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (2a)^2 + (-5b)^2 + (-4c)^2 + 2((2a \times (-5b)) + ((-5b) \times (-4c)) + ((-4c) \times 2a))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 + 2((-10ab) + (20bc) + (-8ca))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 + 2(-10ab) + 2(20bc) + 2(-8ca))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 - 20ab + 40bc - 16ca$ Ans
(iii) $(5x + 3y)^3$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (5x)^3 + 3((5x)^2 \times 3y) + 3(5x \times (3y)^2) + (3y)^3$
$\Rightarrow 125x^3 + 3(25x^2 \times 3y) + 3(5x \times 9y^2) + 27y^3$
$\Rightarrow 125x^3 + 3(75x^2y) + 3(45xy^2) + 27y^3$
$\Rightarrow 125x^3 + 225x^2y + 135xy^2 + 27y^3$ Ans
(iv) $(6a - 7b)^3$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (6a)^3 + 3(6a^2 \times (-7b)) + 3(6a \times (-7b)^2) + (-7b)^3$
$\Rightarrow 216a^3 + 3(36a^2 \times (-7b)) + 3(6a \times (49b^2)) + (-343b^3)$
$\Rightarrow 216a^3 + 3(-252a^2b) + 3(294ab^2) - 343b^3$
$\Rightarrow 216a^3 - 756a^2b + 882ab^2 - 343b^3$ Ans
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (3x)^2 + (-4y)^2 + (5z)^2 + 2((3x \times (-4y)) + ((-4y) \times 5z) + (5z \times 3x))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 + 2((-12xy) + (-20yz) + (15xz))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 + 2(-12xy) + 2(-20yz) + 2(15xz))$
$\Rightarrow 9x^2 + 16y^2 + 25z^2 - 24xy - 40yz + 30xz$ Ans
(ii) $(2a - 5b - 4c)^2$
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (2a)^2 + (-5b)^2 + (-4c)^2 + 2((2a \times (-5b)) + ((-5b) \times (-4c)) + ((-4c) \times 2a))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 + 2((-10ab) + (20bc) + (-8ca))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 + 2(-10ab) + 2(20bc) + 2(-8ca))$
$\Rightarrow 4a^2 + 25b^2 + 16c^2 - 20ab + 40bc - 16ca$ Ans
(iii) $(5x + 3y)^3$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (5x)^3 + 3((5x)^2 \times 3y) + 3(5x \times (3y)^2) + (3y)^3$
$\Rightarrow 125x^3 + 3(25x^2 \times 3y) + 3(5x \times 9y^2) + 27y^3$
$\Rightarrow 125x^3 + 3(75x^2y) + 3(45xy^2) + 27y^3$
$\Rightarrow 125x^3 + 225x^2y + 135xy^2 + 27y^3$ Ans
(iv) $(6a - 7b)^3$
$\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (6a)^3 + 3(6a^2 \times (-7b)) + 3(6a \times (-7b)^2) + (-7b)^3$
$\Rightarrow 216a^3 + 3(36a^2 \times (-7b)) + 3(6a \times (49b^2)) + (-343b^3)$
$\Rightarrow 216a^3 + 3(-252a^2b) + 3(294ab^2) - 343b^3$
$\Rightarrow 216a^3 - 756a^2b + 882ab^2 - 343b^3$ Ans
Q38. If $a + b + c = 9$ and $ab + bc + ca = 15$, find $a^2 + b^2 + c^2$.
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given $a + b + c = 9$ and $ab + bc + ca = 15$ we have:
$\therefore (9)^2 = a^2 + b^2 + c^2 + 2(15)$
$\Rightarrow 81 = a^2 + b^2 + c^2 + 30$
$\Rightarrow 81 - 30 = a^2 + b^2 + c^2$
$\Rightarrow a^2 + b^2 + c^2 = 51$ Ans
Given $a + b + c = 9$ and $ab + bc + ca = 15$ we have:
$\therefore (9)^2 = a^2 + b^2 + c^2 + 2(15)$
$\Rightarrow 81 = a^2 + b^2 + c^2 + 30$
$\Rightarrow 81 - 30 = a^2 + b^2 + c^2$
$\Rightarrow a^2 + b^2 + c^2 = 51$ Ans
Q39. If $a + b + c = 11$ and $a^2 + b^2 + c^2 = 81$, find $ab + bc + ca$.
Show Answer
$\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given $a + b + c = 11$ and $a^2 + b^2 + c^2 = 81$ we have:
$\therefore (11)^2 = 81 + 2(ab + bc + ca)$
$\Rightarrow 121 = 81 + 2(ab + bc + ca)$
$\Rightarrow 121 - 81 = 2(ab + bc + ca)$
$\Rightarrow 2(ab + bc + ca) = 40$
$\Rightarrow ab + bc + ca = \dfrac{40}{2} = 20$ Ans
Given $a + b + c = 11$ and $a^2 + b^2 + c^2 = 81$ we have:
$\therefore (11)^2 = 81 + 2(ab + bc + ca)$
$\Rightarrow 121 = 81 + 2(ab + bc + ca)$
$\Rightarrow 121 - 81 = 2(ab + bc + ca)$
$\Rightarrow 2(ab + bc + ca) = 40$
$\Rightarrow ab + bc + ca = \dfrac{40}{2} = 20$ Ans
Q40. If $3x - 4y = 5$ and $xy = 3$, find $27x^3 - 64y^3$.
Show Answer
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (3x - 4y)^3 = 27x^3 - 64y^3 - 3xy(3x - 4y)$
Given $3x - 4y = 5$ and $xy = 3$ where $x = 3$ and $y = 4$ we have:
$\Rightarrow (5)^3 = 27x^3 - 64y^3 - 3(3(3 \times 4))(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 3(3(12))(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 3(36)(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 108(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 540$
$\Rightarrow 125 + 540 = 27x^3 - 64y^3$
$\Rightarrow 27x^3 - 64y^3 = 665$ Ans
$\therefore (3x - 4y)^3 = 27x^3 - 64y^3 - 3xy(3x - 4y)$
Given $3x - 4y = 5$ and $xy = 3$ where $x = 3$ and $y = 4$ we have:
$\Rightarrow (5)^3 = 27x^3 - 64y^3 - 3(3(3 \times 4))(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 3(3(12))(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 3(36)(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 108(5)$
$\Rightarrow 125 = 27x^3 - 64y^3 - 540$
$\Rightarrow 125 + 540 = 27x^3 - 64y^3$
$\Rightarrow 27x^3 - 64y^3 = 665$ Ans
Q41. If $a + b = 8$ and $ab = 15$, find $a^3 - b^3$.
Show Answer
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
Given $a + b = 8$ and $ab = 15$ we have:
$\therefore (8)^3 = a^3 + b^3 + 3(15)(8)$
$\Rightarrow 512 = a^3 + b^3 + 45(8)$
$\Rightarrow 512 = a^3 + b^3 + 360$
$\Rightarrow 512 - 360 = a^3 + b^3$
$\Rightarrow 27x^3 - 64y^3 = 152$ Ans
Given $a + b = 8$ and $ab = 15$ we have:
$\therefore (8)^3 = a^3 + b^3 + 3(15)(8)$
$\Rightarrow 512 = a^3 + b^3 + 45(8)$
$\Rightarrow 512 = a^3 + b^3 + 360$
$\Rightarrow 512 - 360 = a^3 + b^3$
$\Rightarrow 27x^3 - 64y^3 = 152$ Ans
Q42. If $3x + 2y = 9$ and $xy = 3$, find $27x^3 + 8y^3$.
Show Answer
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
$\therefore (3x + 2y)^3 = 27x^3 + 8y^3 + 3xy(3x + 2y)$
Given $3x + 2y = 9$ and $xy = 3$ where $x = 3$ and $y = 2$ we have:
$\Rightarrow (9)^3 = 27x^3 + 8y^3 + 3(3(3 \times 2))(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 3(3(6))(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 3(18)(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 54(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 486$
$\Rightarrow 729 - 486 = 27x^3 + 8y^3$
$\Rightarrow 27x^3 + 8y^3 = 243$ Ans
$\therefore (3x + 2y)^3 = 27x^3 + 8y^3 + 3xy(3x + 2y)$
Given $3x + 2y = 9$ and $xy = 3$ where $x = 3$ and $y = 2$ we have:
$\Rightarrow (9)^3 = 27x^3 + 8y^3 + 3(3(3 \times 2))(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 3(3(6))(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 3(18)(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 54(9)$
$\Rightarrow 729 = 27x^3 + 8y^3 + 486$
$\Rightarrow 729 - 486 = 27x^3 + 8y^3$
$\Rightarrow 27x^3 + 8y^3 = 243$ Ans
Q43. If $5x - 4y = 7$ and $xy = 8$, find $125x^3 - 64y^3$.
Show Answer
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (5x - 4y)^3 = 125x^3 - 64y^3 - 3xy(5x - 4y)$
Given $5x - 4y = 7$ and $xy = 8$ where $x = 5$ and $y = 4$ we have:
$\Rightarrow (7)^3 = 125x^3 - 64y^3 - 3(8(5 \times 4))(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3(8(20))(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3(160)(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 480(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3360$
$\Rightarrow 343 + 3360 = 125x^3 - 64y^3$
$\Rightarrow 125x^3 - 64y^3 = 3703$ Ans
$\therefore (5x - 4y)^3 = 125x^3 - 64y^3 - 3xy(5x - 4y)$
Given $5x - 4y = 7$ and $xy = 8$ where $x = 5$ and $y = 4$ we have:
$\Rightarrow (7)^3 = 125x^3 - 64y^3 - 3(8(5 \times 4))(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3(8(20))(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3(160)(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 480(7)$
$\Rightarrow 343 = 125x^3 - 64y^3 - 3360$
$\Rightarrow 343 + 3360 = 125x^3 - 64y^3$
$\Rightarrow 125x^3 - 64y^3 = 3703$ Ans
Q44. The difference between two numbers is 5 and their product is 14. Find the difference between their cubes.
Show Answer
Let two numbers be = $x$ and $y$
Given $x - y = 5$ and $xy = 14$
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (x - y)^3 = x^3 - y^3 - 3xy(x - y)$
$\Rightarrow (5)^3 = x^3 - y^3 - 3(14)(5)$
$\Rightarrow 125 = x^3 - y^3 - 42(5)$
$\Rightarrow 125 = x^3 - y^3 - 210$
$\Rightarrow 125 + 210 = x^3 - y^3$
$\Rightarrow x^3 - y^3 = 335$ Ans
Given $x - y = 5$ and $xy = 14$
$\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)$
$\therefore (x - y)^3 = x^3 - y^3 - 3xy(x - y)$
$\Rightarrow (5)^3 = x^3 - y^3 - 3(14)(5)$
$\Rightarrow 125 = x^3 - y^3 - 42(5)$
$\Rightarrow 125 = x^3 - y^3 - 210$
$\Rightarrow 125 + 210 = x^3 - y^3$
$\Rightarrow x^3 - y^3 = 335$ Ans
Q45. Find the square of:
(i) $2a + b$
(ii) $3a + 7b$
(iii) $3a - 4b$
(iv) $5a - b^2$
(v) $a^2 - \dfrac{b}{2}$
(vi) $\dfrac{3a}{2b} - \dfrac{2b}{3a}$
(vii) $a + 2b + c$
(viii) $a + \dfrac{1}{a} - 1$
(ix) $2x - b - 3c$
(x) $\dfrac{2a}{3} - 1 + \dfrac{3}{2a}$
Show Answer
(i) $\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (2a)^2 + 2(2a \times b) + (b)^2$
$\Rightarrow 4a^2 + 2(2ab) + b^2$
$\Rightarrow 4a^2 + 4ab + b^2$ Ans
(ii) $\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (3a)^2 + 2(3a \times 7b) + (7b)^2$
$\Rightarrow 9a^2 + 2(21ab) + 49b^2$
$\Rightarrow 9a^2 + 42ab + 49b^2$ Ans
(iii) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (3a)^2 - 2(3a \times 4b) + (4b)^2$
$\Rightarrow 9a^2 - 2(12ab) + 16b^2$
$\Rightarrow 9a^2 - 24ab + 16b^2$ Ans
(iv) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (5a)^2 - 2(5a \times b^2) + (b^2)^2$
$\Rightarrow 25a^2 - 2(5ab^2) + b^4$
$\Rightarrow 25a^2 - 10ab^2 + b^4$ Ans
(v) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2)^2 - 2(a^2 \times \dfrac{b}{2}) + (\dfrac{b}{2})^2$
$\Rightarrow a^4 - 2(\dfrac{a^2b}{2}) + \dfrac{b^2}{4}$
$\Rightarrow a^4 - a^2b + \dfrac{b^2}{4}$ Ans
(vi) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (\dfrac{3a}{2b})^2 - 2(\dfrac{3a}{2b} \times \dfrac{2b}{3a}) + (\dfrac{2b}{3a})^2$
$\Rightarrow \dfrac{9a^2}{4b^2} - 2(1) + \dfrac{4b^2}{9a^2}$
$\Rightarrow \dfrac{9a^2}{4b^2} - 2 + \dfrac{4b^2}{9a^2}$ Ans
(vii) $\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (a)^2 + (2b)^2 + (c)^2 + 2((a \times 2b) + (2b \times c) + ca)$
$\Rightarrow a^2 + 4b^2 + c^2 + 2(2ab + 2bc + ca)$
$\Rightarrow a^2 + 4b^2 + c^2 + 2(2ab) + 2(2bc) + 2ca$
$\Rightarrow a^2 + 4b^2 + c^2 + 4ab + 4bc + 2ca$ Ans
(viii) $\because (a + b - c)^2 = a^2 + b^2 + c^2 + 2(ab - bc - ca)$
$\therefore (a)^2 + (\dfrac{1}{a})^2 + (1)^2 + 2((a \times \dfrac{1}{a}) - (\dfrac{1}{a} \times 1) - (1 \times a))$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2((1) - \dfrac{1}{a} - a)$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2(1) - 2(\dfrac{1}{a}) - 2(a)$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2 - \dfrac{2}{a} - 2a$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 3 - \dfrac{2}{a} - 2a$ Ans
(ix) $\because (a - b - c)^2 = a^2 + b^2 + c^2 - 2(ab + bc - ca)$
$\therefore (2a)^2 + (b)^2 + (3c)^2 - 2((2a \times b) + (b \times 3c) - (3c \times 2a))$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 2(4ab + 3bc - 6ac)$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 2(4ab) + 2(3bc) - 2(6ac)$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 8ab + 6bc - 12ac$ Ans
(x) $\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2(ab - bc + ca)$
$\therefore (\dfrac{2a}{3})^2 + (1)^2 + (\dfrac{3}{2a})^2 - 2((\dfrac{2a}{3} \times 1) - (1 \times \dfrac{3}{2a}) + (\dfrac{3}{2a} \times \dfrac{2a}{3}))$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - 2(\dfrac{2a}{3} - \dfrac{3}{2a} + \dfrac{3}{2a})$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - 2(\dfrac{2a}{3}) - 2(\dfrac{3}{2a}) + 2(1)$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a} + 2$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a} + 2$
$\Rightarrow \dfrac{4a^2}{9} + 3 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a}$ Ans
$\therefore (2a)^2 + 2(2a \times b) + (b)^2$
$\Rightarrow 4a^2 + 2(2ab) + b^2$
$\Rightarrow 4a^2 + 4ab + b^2$ Ans
(ii) $\because (a + b)^2 = a^2 + 2ab + b^2$
$\therefore (3a)^2 + 2(3a \times 7b) + (7b)^2$
$\Rightarrow 9a^2 + 2(21ab) + 49b^2$
$\Rightarrow 9a^2 + 42ab + 49b^2$ Ans
(iii) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (3a)^2 - 2(3a \times 4b) + (4b)^2$
$\Rightarrow 9a^2 - 2(12ab) + 16b^2$
$\Rightarrow 9a^2 - 24ab + 16b^2$ Ans
(iv) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (5a)^2 - 2(5a \times b^2) + (b^2)^2$
$\Rightarrow 25a^2 - 2(5ab^2) + b^4$
$\Rightarrow 25a^2 - 10ab^2 + b^4$ Ans
(v) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (a^2)^2 - 2(a^2 \times \dfrac{b}{2}) + (\dfrac{b}{2})^2$
$\Rightarrow a^4 - 2(\dfrac{a^2b}{2}) + \dfrac{b^2}{4}$
$\Rightarrow a^4 - a^2b + \dfrac{b^2}{4}$ Ans
(vi) $\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore (\dfrac{3a}{2b})^2 - 2(\dfrac{3a}{2b} \times \dfrac{2b}{3a}) + (\dfrac{2b}{3a})^2$
$\Rightarrow \dfrac{9a^2}{4b^2} - 2(1) + \dfrac{4b^2}{9a^2}$
$\Rightarrow \dfrac{9a^2}{4b^2} - 2 + \dfrac{4b^2}{9a^2}$ Ans
(vii) $\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
$\therefore (a)^2 + (2b)^2 + (c)^2 + 2((a \times 2b) + (2b \times c) + ca)$
$\Rightarrow a^2 + 4b^2 + c^2 + 2(2ab + 2bc + ca)$
$\Rightarrow a^2 + 4b^2 + c^2 + 2(2ab) + 2(2bc) + 2ca$
$\Rightarrow a^2 + 4b^2 + c^2 + 4ab + 4bc + 2ca$ Ans
(viii) $\because (a + b - c)^2 = a^2 + b^2 + c^2 + 2(ab - bc - ca)$
$\therefore (a)^2 + (\dfrac{1}{a})^2 + (1)^2 + 2((a \times \dfrac{1}{a}) - (\dfrac{1}{a} \times 1) - (1 \times a))$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2((1) - \dfrac{1}{a} - a)$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2(1) - 2(\dfrac{1}{a}) - 2(a)$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 1 + 2 - \dfrac{2}{a} - 2a$
$\Rightarrow a^2 + \dfrac{1}{a^2} + 3 - \dfrac{2}{a} - 2a$ Ans
(ix) $\because (a - b - c)^2 = a^2 + b^2 + c^2 - 2(ab + bc - ca)$
$\therefore (2a)^2 + (b)^2 + (3c)^2 - 2((2a \times b) + (b \times 3c) - (3c \times 2a))$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 2(4ab + 3bc - 6ac)$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 2(4ab) + 2(3bc) - 2(6ac)$
$\Rightarrow 4a^2 + b^2 + 9c^2 - 8ab + 6bc - 12ac$ Ans
(x) $\because (a - b + c)^2 = a^2 + b^2 + c^2 - 2(ab - bc + ca)$
$\therefore (\dfrac{2a}{3})^2 + (1)^2 + (\dfrac{3}{2a})^2 - 2((\dfrac{2a}{3} \times 1) - (1 \times \dfrac{3}{2a}) + (\dfrac{3}{2a} \times \dfrac{2a}{3}))$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - 2(\dfrac{2a}{3} - \dfrac{3}{2a} + \dfrac{3}{2a})$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - 2(\dfrac{2a}{3}) - 2(\dfrac{3}{2a}) + 2(1)$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a} + 2$
$\Rightarrow \dfrac{4a^2}{9} + 1 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a} + 2$
$\Rightarrow \dfrac{4a^2}{9} + 3 + \dfrac{9}{4a^2} - \dfrac{4a}{3} - \dfrac{3}{a}$ Ans