Q61. If $a^2 - 5a - 1 = 0$, find:
(i) $a - \dfrac{1}{a}$
(ii) $a + \dfrac{1}{a}$
(iii) $a^2 - \dfrac{1}{a^2}$
Show Answer
(i) Upon dividing each term by $a$ we have:
$\Rightarrow \dfrac{a^2}{a} - \dfrac{5a}{a} - \dfrac{1}{a} = 0$
$\Rightarrow a - 5 - \dfrac{1}{a} = 0$
$\Rightarrow a - \dfrac{1}{a} = 0 + 5$
$\Rightarrow a - \dfrac{1}{a} = 5$ Ans
(ii) $\because (a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4$
From (i) we have $a - \dfrac{1}{a} = 5$
$\therefore (a + \dfrac{1}{a})^2 - (5)^2 = 4$
$\Rightarrow (a + \dfrac{1}{a})^2 - 25 = 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 4 + 25$
$\Rightarrow (a + \dfrac{1}{a})^2 = 29$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{29}$ Ans
(iii) $\because a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
From (i) we have $a - \dfrac{1}{a} = 5$
From (ii) we have $a + \dfrac{1}{a} = \pm\sqrt{29}$
$\therefore a^2 - b^2 = (5)(\sqrt{29})$
$\Rightarrow a^2 - b^2 = 5 \times \sqrt{29} = \pm\sqrt[5]{29}$ Ans
$\Rightarrow \dfrac{a^2}{a} - \dfrac{5a}{a} - \dfrac{1}{a} = 0$
$\Rightarrow a - 5 - \dfrac{1}{a} = 0$
$\Rightarrow a - \dfrac{1}{a} = 0 + 5$
$\Rightarrow a - \dfrac{1}{a} = 5$ Ans
(ii) $\because (a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4$
From (i) we have $a - \dfrac{1}{a} = 5$
$\therefore (a + \dfrac{1}{a})^2 - (5)^2 = 4$
$\Rightarrow (a + \dfrac{1}{a})^2 - 25 = 4$
$\Rightarrow (a + \dfrac{1}{a})^2 = 4 + 25$
$\Rightarrow (a + \dfrac{1}{a})^2 = 29$
$\Rightarrow a + \dfrac{1}{a} = \pm\sqrt{29}$ Ans
(iii) $\because a^2 - \dfrac{1}{a^2} = (a + \dfrac{1}{a})(a - \dfrac{1}{a})$
From (i) we have $a - \dfrac{1}{a} = 5$
From (ii) we have $a + \dfrac{1}{a} = \pm\sqrt{29}$
$\therefore a^2 - b^2 = (5)(\sqrt{29})$
$\Rightarrow a^2 - b^2 = 5 \times \sqrt{29} = \pm\sqrt[5]{29}$ Ans
Q62. If $3x + 4y = 16$ and $xy = 4$, find the value of $9x^2 + 16y^2$
Show Answer
$\because (a + b)^2 = a^2 + b^2 + 2ab$
$\therefore (3x + 4y)^2 = 9x^2 + 16y^2 + 2xy$
Here we have: $x = 3, y = 4$ and $xy = 4$ so we have:
$\Rightarrow (16)^2 = 9x^2 + 16y^2 + 2(3 \times 4 \times 4)$
$\Rightarrow 256 = 9x^2 + 16y^2 + 2(48)$
$\Rightarrow 256 = 9x^2 + 16y^2 + 96$
$\Rightarrow 9x^2 + 16y^2 = 256 - 96$
$\Rightarrow 9x^2 + 16y^2 = 160$ Ans
$\therefore (3x + 4y)^2 = 9x^2 + 16y^2 + 2xy$
Here we have: $x = 3, y = 4$ and $xy = 4$ so we have:
$\Rightarrow (16)^2 = 9x^2 + 16y^2 + 2(3 \times 4 \times 4)$
$\Rightarrow 256 = 9x^2 + 16y^2 + 2(48)$
$\Rightarrow 256 = 9x^2 + 16y^2 + 96$
$\Rightarrow 9x^2 + 16y^2 = 256 - 96$
$\Rightarrow 9x^2 + 16y^2 = 160$ Ans
Q63. Find the cube of:
(i) $3a - 2b$
(ii) $5a + 3b$
(iii) $2a + \dfrac{1}{2a}$
(iv) $3a - \dfrac{1}{a}$
Show Answer
(i) $\because (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$\therefore (3a - 2b)^3 = (3a)^3 - 3((3a)^2 \times 2b) + 3(3a \times (2b)^2) - (2b)^3$
$\Rightarrow 27a^3 - 3(9a^2 \times 2b) + 3(3a \times 4b^2) - 8b^3$
$\Rightarrow 27a^3 - 3(18a^2b) + 3(12ab^2) - 8b^3$
$\Rightarrow 27a^3 - 54a^2b + 36ab^2 - 8b^3$ Ans
(ii) $\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (5a + 3b)^3 = (5a)^3 + 3((5a)^2 \times 3b) + 3(5a \times (3b)^2) + (3b)^3$
$\Rightarrow 125a^3 + 3(25a^2 \times 3b) + 3(5a \times 9b^2) + 27b^3$
$\Rightarrow 125a^3 + 3(75a^2b) + 3(45ab^2) + 27b^3$
$\Rightarrow 125a^3 + 225a^2b + 135ab^2 + 27b^3$ Ans
(iii) $\because (a + \dfrac{1}{a})^3 = a^3 + \dfrac{1}{a^3} + 3(a + \dfrac{1}{a})$
$\therefore (2a + \dfrac{1}{2a})^3 = (2a)^3 + (\dfrac{1}{2a})^3 + 3(2a + \dfrac{1}{2a})$
$\Rightarrow 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}$
$\therefore 8a^3 + 6a + \dfrac{3}{2a} + \dfrac{1}{8a^3}$ Ans
(iv) $\because (a - \dfrac{1}{a})^3 = a^3 - \dfrac{1}{a^3} - 3(a - \dfrac{1}{a})$
$\therefore (3a - \dfrac{1}{a})^3 = (3a)^3 - (\dfrac{1}{a})^3 - 3(3a - \dfrac{1}{a})$
$\Rightarrow 27a^3 - \dfrac{1}{a^3} - 9a + \dfrac{3}{a}$
$\therefore 27a^3 - 9a + \dfrac{3}{a} - \dfrac{1}{a^3}$ Ans
$\therefore (3a - 2b)^3 = (3a)^3 - 3((3a)^2 \times 2b) + 3(3a \times (2b)^2) - (2b)^3$
$\Rightarrow 27a^3 - 3(9a^2 \times 2b) + 3(3a \times 4b^2) - 8b^3$
$\Rightarrow 27a^3 - 3(18a^2b) + 3(12ab^2) - 8b^3$
$\Rightarrow 27a^3 - 54a^2b + 36ab^2 - 8b^3$ Ans
(ii) $\because (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$\therefore (5a + 3b)^3 = (5a)^3 + 3((5a)^2 \times 3b) + 3(5a \times (3b)^2) + (3b)^3$
$\Rightarrow 125a^3 + 3(25a^2 \times 3b) + 3(5a \times 9b^2) + 27b^3$
$\Rightarrow 125a^3 + 3(75a^2b) + 3(45ab^2) + 27b^3$
$\Rightarrow 125a^3 + 225a^2b + 135ab^2 + 27b^3$ Ans
(iii) $\because (a + \dfrac{1}{a})^3 = a^3 + \dfrac{1}{a^3} + 3(a + \dfrac{1}{a})$
$\therefore (2a + \dfrac{1}{2a})^3 = (2a)^3 + (\dfrac{1}{2a})^3 + 3(2a + \dfrac{1}{2a})$
$\Rightarrow 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}$
$\therefore 8a^3 + 6a + \dfrac{3}{2a} + \dfrac{1}{8a^3}$ Ans
(iv) $\because (a - \dfrac{1}{a})^3 = a^3 - \dfrac{1}{a^3} - 3(a - \dfrac{1}{a})$
$\therefore (3a - \dfrac{1}{a})^3 = (3a)^3 - (\dfrac{1}{a})^3 - 3(3a - \dfrac{1}{a})$
$\Rightarrow 27a^3 - \dfrac{1}{a^3} - 9a + \dfrac{3}{a}$
$\therefore 27a^3 - 9a + \dfrac{3}{a} - \dfrac{1}{a^3}$ Ans
Q64. If $a^2 + \dfrac{1}{a^2} = 47$, find:
(i) $a + \dfrac{1}{a}$
(ii) $a^3 + \dfrac{1}{a^3}$
Show Answer
(i) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a^2 + \dfrac{1}{a^2} = 47$ we have:
$\therefore (a + \dfrac{1}{a})^2 = 47 + 2$
$\Rightarrow (a + \dfrac{1}{a})^2 = 49$
$\Rightarrow a + \dfrac{1}{a} = \sqrt{49}$
$\Rightarrow a + \dfrac{1}{a} = 7$ Ans
(ii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
From (i) we have: $a + \dfrac{1}{a} = 7$
$\therefore a^3 + \dfrac{1}{a^3} = (7)^3 - 3(7)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 343 - 21$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 322$ Ans
Given $a^2 + \dfrac{1}{a^2} = 47$ we have:
$\therefore (a + \dfrac{1}{a})^2 = 47 + 2$
$\Rightarrow (a + \dfrac{1}{a})^2 = 49$
$\Rightarrow a + \dfrac{1}{a} = \sqrt{49}$
$\Rightarrow a + \dfrac{1}{a} = 7$ Ans
(ii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
From (i) we have: $a + \dfrac{1}{a} = 7$
$\therefore a^3 + \dfrac{1}{a^3} = (7)^3 - 3(7)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 343 - 21$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 322$ Ans
Q65. If $a^2 + \dfrac{1}{a^2} = 18$, find:
(i) $a - \dfrac{1}{a}$
(ii) $a^3 - \dfrac{1}{a^3}$
Show Answer
(i) $\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $a^2 + \dfrac{1}{a^2} = 18$ we have:
$\therefore (a - \dfrac{1}{a})^2 = 18 - 2$
$\Rightarrow (a - \dfrac{1}{a})^2 = 16$
$\Rightarrow a - \dfrac{1}{a} = \sqrt{16}$
$\Rightarrow a - \dfrac{1}{a} = 4$ Ans
(ii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
From (i) we have: $a - \dfrac{1}{a} = 4$
$\therefore a^3 - \dfrac{1}{a^3} = (4)^3 + 3(4)$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 64 + 12$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 76$ Ans
Given $a^2 + \dfrac{1}{a^2} = 18$ we have:
$\therefore (a - \dfrac{1}{a})^2 = 18 - 2$
$\Rightarrow (a - \dfrac{1}{a})^2 = 16$
$\Rightarrow a - \dfrac{1}{a} = \sqrt{16}$
$\Rightarrow a - \dfrac{1}{a} = 4$ Ans
(ii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
From (i) we have: $a - \dfrac{1}{a} = 4$
$\therefore a^3 - \dfrac{1}{a^3} = (4)^3 + 3(4)$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 64 + 12$
$\Rightarrow a^3 - \dfrac{1}{a^3} = 76$ Ans
Q66. If $a + \dfrac{1}{a} = p$, then show that $a^3 + \dfrac{1}{a^3} = p(p^2 - 3)$:
Show Answer
$\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $a + \dfrac{1}{a} = p$ we have:
$\therefore a^3 + \dfrac{1}{a^3} = (p)^3 - 3(p)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 3p^3 - 3p$
$\Rightarrow a^3 + \dfrac{1}{a^3} = p(3p^2 - 3)$ Ans
Given $a + \dfrac{1}{a} = p$ we have:
$\therefore a^3 + \dfrac{1}{a^3} = (p)^3 - 3(p)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 3p^3 - 3p$
$\Rightarrow a^3 + \dfrac{1}{a^3} = p(3p^2 - 3)$ Ans
Q67. If $a + 2b = 5$, then show that $a^3 + 8b^3 + 30ab = 125$:
Show Answer
$\because a^3 + b^3 = (a + b)^3 - 3ab(a + b)$
Given $a + 2b = 5$ we have:
$\therefore (a)^3 + (2b)^3 = (a + 2b)^3 - 3(a \times 2b)(a + 2b)$
$\Rightarrow a^3 + 8b^3 = (5)^3 - 3(2ab)(5)$
$\Rightarrow a^3 + 8b^3 = 125 - 6ab(5)$
$\Rightarrow a^3 + 8b^3 = 125 - 30ab$
$\Rightarrow a^3 + 8b^3 + 30ab = 125$ Proved
Given $a + 2b = 5$ we have:
$\therefore (a)^3 + (2b)^3 = (a + 2b)^3 - 3(a \times 2b)(a + 2b)$
$\Rightarrow a^3 + 8b^3 = (5)^3 - 3(2ab)(5)$
$\Rightarrow a^3 + 8b^3 = 125 - 6ab(5)$
$\Rightarrow a^3 + 8b^3 = 125 - 30ab$
$\Rightarrow a^3 + 8b^3 + 30ab = 125$ Proved
Q68. If $(a + \dfrac{1}{a})^2 = 3$, then show that $a^3 + \dfrac{1}{a^3} = 0$:
Show Answer
Given: $(a + \dfrac{1}{a})^2 = 3$
$\therefore a + \dfrac{1}{a} = \sqrt{3}$
$\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
$\therefore a^3 + \dfrac{1}{a^3} = (\sqrt{3})^3 - 3(a + \dfrac{1}{a})$
$\Rightarrow a^3 + \dfrac{1}{a^3} = \sqrt[3]{3} - \sqrt[3]{3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = \sqrt[3]{3} - \sqrt[3]{3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 0$ Proved
$\therefore a + \dfrac{1}{a} = \sqrt{3}$
$\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
$\therefore a^3 + \dfrac{1}{a^3} = (\sqrt{3})^3 - 3(a + \dfrac{1}{a})$
$\Rightarrow a^3 + \dfrac{1}{a^3} = \sqrt[3]{3} - \sqrt[3]{3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = \sqrt[3]{3} - \sqrt[3]{3}$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 0$ Proved
Q69. If $a + 2b + c = 0$, then show that $a^3 + 8b^3 + c^3 = 6abc$:
Show Answer
$\because a + 2b + c = 0$
$\therefore a + 2b = -c$
Given: $a^3 + 8b^3 + c^3 = 6abc$
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
When applying the above formula on $a + 2b = -c$, we have:
$\therefore (a)^3 + (2b)^3 + 3(a \times 2b)(a + 2b) = (-c)^3$
$\Rightarrow a^3 + 8b^3 + 3(2ab)(-c) = -c^3$
$\Rightarrow a^3 + 8b^3 + 6ab(-c) = -c^3$
$\Rightarrow a^3 + 8b^3 - 6abc = -c^3$
$\therefore a^3 + 8b^3 - c^3 = 6abc$ Proved
$\therefore a + 2b = -c$
Given: $a^3 + 8b^3 + c^3 = 6abc$
$\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$
When applying the above formula on $a + 2b = -c$, we have:
$\therefore (a)^3 + (2b)^3 + 3(a \times 2b)(a + 2b) = (-c)^3$
$\Rightarrow a^3 + 8b^3 + 3(2ab)(-c) = -c^3$
$\Rightarrow a^3 + 8b^3 + 6ab(-c) = -c^3$
$\Rightarrow a^3 + 8b^3 - 6abc = -c^3$
$\therefore a^3 + 8b^3 - c^3 = 6abc$ Proved
Q70. If $a + \dfrac{1}{a} = 4$, find:
(i) $a^2 + \dfrac{1}{a^2}$
(ii) $a^4 + \dfrac{1}{a^4}$
(iii) $a^3 + \dfrac{1}{a^3}$
Show Answer
(i) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a + \dfrac{1}{a} = 4$:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 16 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 14$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 14$:
$\therefore (14)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 196 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 196 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 194$ Ans
(iii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $a + \dfrac{1}{a} = 4$:
$\therefore a^3 + \dfrac{1}{a^3} = (4)^3 - 3(4)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 64 - 12$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 52$ Ans
Given $a + \dfrac{1}{a} = 4$:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 16 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 14$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 14$:
$\therefore (14)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 196 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 196 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 194$ Ans
(iii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $a + \dfrac{1}{a} = 4$:
$\therefore a^3 + \dfrac{1}{a^3} = (4)^3 - 3(4)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 64 - 12$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 52$ Ans
Q71. If $a + \dfrac{1}{a} = 6$, find:
(i) $a^2 + \dfrac{1}{a^2}$
(ii) $a^4 + \dfrac{1}{a^4}$
(iii) $a^3 + \dfrac{1}{a^3}$
Show Answer
(i) $\because (a + \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} + 2$
Given $a + \dfrac{1}{a} = 6$:
$\therefore (6)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 36 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 36 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 34$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 34$:
$\therefore (34)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 1156 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 1156 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 1154$ Ans
(iii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $a + \dfrac{1}{a} = 6$:
$\therefore a^3 + \dfrac{1}{a^3} = (6)^3 - 3(6)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 216 - 18$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 198$ Ans
Given $a + \dfrac{1}{a} = 6$:
$\therefore (6)^2 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 36 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 36 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 34$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 34$:
$\therefore (34)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 1156 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 1156 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 1154$ Ans
(iii) $\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $a + \dfrac{1}{a} = 6$:
$\therefore a^3 + \dfrac{1}{a^3} = (6)^3 - 3(6)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 216 - 18$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 198$ Ans
Q72. If $a - \dfrac{1}{a} = 3$, find:
(i) $a^2 + \dfrac{1}{a^2}$
(ii) $a^3 - \dfrac{1}{a^3}$
Show Answer
(i) $\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $a - \dfrac{1}{a} = 3$:
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 9 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 11$ Ans
(ii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
Given $a - \dfrac{1}{a} = 3$:
$\therefore a^3 + \dfrac{1}{a^3} = (3)^3 + 3(3)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 27 + 9$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 36$ Ans
Given $a - \dfrac{1}{a} = 3$:
$\therefore (3)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 9 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 9 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 11$ Ans
(ii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
Given $a - \dfrac{1}{a} = 3$:
$\therefore a^3 + \dfrac{1}{a^3} = (3)^3 + 3(3)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 27 + 9$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 36$ Ans
Q73. If $a - \dfrac{1}{a} = 4$, find:
(i) $a^2 + \dfrac{1}{a^2}$
(ii) $a^4 + \dfrac{1}{a^4}$
(iii) $a^3 - \dfrac{1}{a^3}$
Show Answer
(i) $\because (a - \dfrac{1}{a})^2 = a^2 + \dfrac{1}{a^2} - 2$
Given $a - \dfrac{1}{a} = 4$:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 18$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 18$:
$\therefore (18)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 324 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 324 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 322$ Ans
(iii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
Given $a - \dfrac{1}{a} = 4$:
$\therefore a^3 + \dfrac{1}{a^3} = (4)^3 + 3(4)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 64 + 12$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 76$ Ans
Given $a - \dfrac{1}{a} = 4$:
$\therefore (4)^2 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 = a^2 + \dfrac{1}{a^2} - 2$
$\Rightarrow 16 + 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 18$ Ans
(ii) $\because (a^2 + \dfrac{1}{a^2})^2 = a^4 + \dfrac{1}{a^4} + 2$
From (i) we have: $a^2 + \dfrac{1}{a^2} = 18$:
$\therefore (18)^2 = a^4 + \dfrac{1}{a^4} + 2$
$\Rightarrow 324 = a^2 + \dfrac{1}{a^2} + 2$
$\Rightarrow 324 - 2 = a^2 + \dfrac{1}{a^2}$
$\Rightarrow a^2 + \dfrac{1}{a^2} = 322$ Ans
(iii) $\because a^3 - \dfrac{1}{a^3} = (a - \dfrac{1}{a})^3 + 3(a - \dfrac{1}{a})$
Given $a - \dfrac{1}{a} = 4$:
$\therefore a^3 + \dfrac{1}{a^3} = (4)^3 + 3(4)$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 64 + 12$
$\Rightarrow a^3 + \dfrac{1}{a^3} = 76$ Ans
Q74. $x + \dfrac{1}{x} = 2$, then show that:
$x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$
Show Answer
For $x^2 + \dfrac{1}{x^2}$:
$\because (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$
Given $x + \dfrac{1}{x} = 2$:
$\therefore (2)^2 = x^2 + \dfrac{1}{x^2} + 2$
$\Rightarrow 4 = x^2 + \dfrac{1}{x^2} + 2$
$\Rightarrow 4 - 2 = x^2 + \dfrac{1}{x^2}$
$\therefore x^2 + \dfrac{1}{x^2} = 2$
For $x^3 + \dfrac{1}{x^3}$:
$\because x^3 + \dfrac{1}{x}^3 = (x + \dfrac{1}{x})^3 - 3(x + \dfrac{1}{x}) $
Given $x + \dfrac{1}{x} = 2$:
$\therefore x^3 + \dfrac{1}{x}^3 = (2)^3 - 3(2)$
$\Rightarrow x^3 + \dfrac{1}{x}^3 = 8 - 6$
$\Rightarrow x^3 + \dfrac{1}{x}^3 = 2$
For $x^4 + \dfrac{1}{x^4}$:
$\because (x^2 + \dfrac{1}{x^2})^2 = x^4 + \dfrac{1}{x^4} + 2$
Here: $x^2 + \dfrac{1}{x^2} = 2$:
$\therefore (2)^2 = x^4 + \dfrac{1}{x^4} + 2$
$\Rightarrow 4 - 2 = x^4 + \dfrac{1}{x^4}$
$\Rightarrow x^4 + \dfrac{1}{x^4} = 2$
Hence $x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$ Proved
$\because (x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$
Given $x + \dfrac{1}{x} = 2$:
$\therefore (2)^2 = x^2 + \dfrac{1}{x^2} + 2$
$\Rightarrow 4 = x^2 + \dfrac{1}{x^2} + 2$
$\Rightarrow 4 - 2 = x^2 + \dfrac{1}{x^2}$
$\therefore x^2 + \dfrac{1}{x^2} = 2$
For $x^3 + \dfrac{1}{x^3}$:
$\because x^3 + \dfrac{1}{x}^3 = (x + \dfrac{1}{x})^3 - 3(x + \dfrac{1}{x}) $
Given $x + \dfrac{1}{x} = 2$:
$\therefore x^3 + \dfrac{1}{x}^3 = (2)^3 - 3(2)$
$\Rightarrow x^3 + \dfrac{1}{x}^3 = 8 - 6$
$\Rightarrow x^3 + \dfrac{1}{x}^3 = 2$
For $x^4 + \dfrac{1}{x^4}$:
$\because (x^2 + \dfrac{1}{x^2})^2 = x^4 + \dfrac{1}{x^4} + 2$
Here: $x^2 + \dfrac{1}{x^2} = 2$:
$\therefore (2)^2 = x^4 + \dfrac{1}{x^4} + 2$
$\Rightarrow 4 - 2 = x^4 + \dfrac{1}{x^4}$
$\Rightarrow x^4 + \dfrac{1}{x^4} = 2$
Hence $x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$ Proved
Q75. If $a + \dfrac{1}{a} = m$ and $a - \dfrac{1}{a} = n$, find the relation between $m$ and $n$.
Show Answer
$\because (a + \dfrac{1}{a})^2 - (a - \dfrac{1}{a})^2 = 4$
Given $a + \dfrac{1}{a} = m$ and $a - \dfrac{1}{a} = n$:
$\therefore (m)^2 - (n)^2 = 4$
$\Rightarrow m^2 - n^2 = 4$ Ans
Given $a + \dfrac{1}{a} = m$ and $a - \dfrac{1}{a} = n$:
$\therefore (m)^2 - (n)^2 = 4$
$\Rightarrow m^2 - n^2 = 4$ Ans
Q76. If $3x + \dfrac{1}{3x} = 8$, find the value of $27x^3 + \dfrac{1}{27x^3}$.
Show Answer
$\because a^3 + \dfrac{1}{a^3} = (a + \dfrac{1}{a})^3 - 3(a + \dfrac{1}{a})$
Given $3x + \dfrac{1}{3x} = 8$:
$\therefore (3x)^3 + \dfrac{1}{(3x)^3} = (8)^3 - 3(8)$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 512 - 24$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 488$ Ans
Given $3x + \dfrac{1}{3x} = 8$:
$\therefore (3x)^3 + \dfrac{1}{(3x)^3} = (8)^3 - 3(8)$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 512 - 24$
$\Rightarrow 27x^3 + \dfrac{1}{27x^3} = 488$ Ans
Q77. If $2x - 3y = 10$ and $xy = 16$, find the value of $8x^3 - 27y^3$.
Show Answer
$\because a^3 - b^3 = (a - b)^3 + 3ab(a - b)$
$\therefore x^3 - y^3 = (x - y)^3 + 3xy(x - y)$
Given $2x - 3y = 10$ and $xy = 16$,
Also, here $x = 2$ and $y = 3$:
$\therefore (2x)^3 - (3y)^3 = (2x - 3y)^3 + 3(2 \times 3 \times 16)(2x - 3y)$
$\Rightarrow 8x^3 - 27y^3 = (10)^3 + 3(96)(10)$
$\Rightarrow 8x^3 - 27y^3 = 1000 + 288(10)$
$\Rightarrow 8x^3 - 27y^3 = 1000 + 2880$
$\Rightarrow 8x^3 - 27y^3 = 3880$ Ans
$\therefore x^3 - y^3 = (x - y)^3 + 3xy(x - y)$
Given $2x - 3y = 10$ and $xy = 16$,
Also, here $x = 2$ and $y = 3$:
$\therefore (2x)^3 - (3y)^3 = (2x - 3y)^3 + 3(2 \times 3 \times 16)(2x - 3y)$
$\Rightarrow 8x^3 - 27y^3 = (10)^3 + 3(96)(10)$
$\Rightarrow 8x^3 - 27y^3 = 1000 + 288(10)$
$\Rightarrow 8x^3 - 27y^3 = 1000 + 2880$
$\Rightarrow 8x^3 - 27y^3 = 3880$ Ans
Q78. If $x^3 + y^3 + z^3 = 3xyz$ and $x + y + z = 0$, find the value of:
$\dfrac{(x + y)^2}{xy} + \dfrac{(y + z)^2}{yz} + \dfrac{(z + x)^2}{zx}$
Show Answer
For numerator, from $x + y + z = 0$ we have:
$\because x + y = -z$
$\therefore (x + y)^2 = (-z)^2 = z$
Similarly
$\because y + z = -x$
$\therefore (y + z)^2 = (-x)^2 = x$
And
$\because x + z = -y$
$\therefore (x + z)^2 = (-y)^2 = y$
For denominator, from $x^3 + y^3 + z^3 = 3xyz$ we have:
Taking the form $3xyz = x^3 + y^3 + z^3$
$\Rightarrow xy = \dfrac{x^3 + y^3 + z^3}{3z}$
$\Rightarrow yz = \dfrac{x^3 + y^3 + z^3}{3x}$
$\Rightarrow zx = \dfrac{x^3 + y^3 + z^3}{3y}$
Given:
$\Rightarrow \dfrac{(x + y)^2}{xy} + \dfrac{(y + z)^2}{yz} + \dfrac{(z + x)^2}{zx}$
We have:
$\Rightarrow \dfrac{(x + y)^2}{\dfrac{x^3 + y^3 + z^3}{3z}} + \dfrac{(y + z)^2}{\dfrac{x^3 + y^3 + z^3}{3x}} + \dfrac{(z + x)^2}{\dfrac{x^3 + y^3 + z^3}{3y}}$
$\Rightarrow \dfrac{(z)^2(3z)}{x^3 + y^3 + z^3} + \dfrac{(x)^2(3x)}{x^3 + y^3 + z^3} + \dfrac{(y)^2(3y)}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3z^3}{x^3 + y^3 + z^3} + \dfrac{3x^3}{x^3 + y^3 + z^3} + \dfrac{3y^3}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3z^3 + 3x^3 + 3y^3}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3(z^3 + x^3 + y^3)}{x^3 + y^3 + z^3} = 3$ Ans
$\because x + y = -z$
$\therefore (x + y)^2 = (-z)^2 = z$
Similarly
$\because y + z = -x$
$\therefore (y + z)^2 = (-x)^2 = x$
And
$\because x + z = -y$
$\therefore (x + z)^2 = (-y)^2 = y$
For denominator, from $x^3 + y^3 + z^3 = 3xyz$ we have:
Taking the form $3xyz = x^3 + y^3 + z^3$
$\Rightarrow xy = \dfrac{x^3 + y^3 + z^3}{3z}$
$\Rightarrow yz = \dfrac{x^3 + y^3 + z^3}{3x}$
$\Rightarrow zx = \dfrac{x^3 + y^3 + z^3}{3y}$
Given:
$\Rightarrow \dfrac{(x + y)^2}{xy} + \dfrac{(y + z)^2}{yz} + \dfrac{(z + x)^2}{zx}$
We have:
$\Rightarrow \dfrac{(x + y)^2}{\dfrac{x^3 + y^3 + z^3}{3z}} + \dfrac{(y + z)^2}{\dfrac{x^3 + y^3 + z^3}{3x}} + \dfrac{(z + x)^2}{\dfrac{x^3 + y^3 + z^3}{3y}}$
$\Rightarrow \dfrac{(z)^2(3z)}{x^3 + y^3 + z^3} + \dfrac{(x)^2(3x)}{x^3 + y^3 + z^3} + \dfrac{(y)^2(3y)}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3z^3}{x^3 + y^3 + z^3} + \dfrac{3x^3}{x^3 + y^3 + z^3} + \dfrac{3y^3}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3z^3 + 3x^3 + 3y^3}{x^3 + y^3 + z^3}$
$\Rightarrow \dfrac{3(z^3 + x^3 + y^3)}{x^3 + y^3 + z^3} = 3$ Ans