If $a^2 + \dfrac{1}{a^2} = 18$ and a ≠ 0; find :

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

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

(i) By formula,

$\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 18 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 16 \\[1em] \Rightarrow a - \dfrac{1}{a} = \sqrt{16} \\[1em] \Rightarrow a - \dfrac{1}{a} = \pm 4.$

Hence, $a - \dfrac{1}{a} = \pm 4$.

(ii) By formula,

$\Big(a - \dfrac{1}{a}\Big)^3 = a^3 - \dfrac{1}{a^3} -3\Big(a - \dfrac{1}{a}\Big)$

Substituting $a - \dfrac{1}{a} = 4$ we get :

$\Rightarrow 4^3 = a^3 - \dfrac{1}{a^3} - 3 \times 4 \\[1em] \Rightarrow 64 = a^3 - \dfrac{1}{a^3} - 12 \\[1em] \Rightarrow 64 + 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = 76.$

Substituting $a - \dfrac{1}{a} = -4$ we get :

$\Rightarrow (-4)^3 = a^3 - \dfrac{1}{a^3} - 3 \times -4 \\[1em] \Rightarrow -64 = a^3 - \dfrac{1}{a^3} + 12 \\[1em] \Rightarrow -64 - 12 = a^3 - \dfrac{1}{a^3} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} = -76.$

Hence, $a^3 - \dfrac{1}{a^3} = \pm 76$.