If $a^2 + \dfrac{1}{a^2} = 47$ and a ≠ 0; find :

(i) $a + \dfrac{1}{a}$

(ii) $a^3 + \dfrac{1}{a^3}$

(i) By formula,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 47 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 49 \\[1em] \Rightarrow a + \dfrac{1}{a} = \sqrt{49} \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm 7.$

Hence, $a + \dfrac{1}{a} = \pm 7.$

(ii) By formula,

$\Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$ ……..(1)

Substituting $a + \dfrac{1}{a} = 7$ in equation (1), we get :

$\Rightarrow 7^3 = a^3 + \dfrac{1}{a^3} + 3 \times 7 \\[1em] \Rightarrow 343 = a^3 + \dfrac{1}{a^3} + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 343 - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 322.$

Substituting $a + \dfrac{1}{a} = -7$ in equation (1), we get :

$\Rightarrow (-7)^3 = a^3 + \dfrac{1}{a^3} + 3 \times -7 \\[1em] \Rightarrow -343 = a^3 + \dfrac{1}{a^3} - 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -343 + 21 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -322.$

Hence, $a^3 + \dfrac{1}{a^3} = \pm 322.$