If a ≠ 0 and $a - \dfrac{1}{a} = 4$; find :

(i) $a^2 + \dfrac{1}{a^2}$

(ii) $a^4 + \dfrac{1}{a^4}$

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

(i) By formula,

$\Rightarrow a^2 + \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)^2 + 2$

Substituting values we get :

$\Rightarrow a^2 + \dfrac{1}{a^2} = 4^2 + 2 \\[1em] = 16 + 2 \\[1em] = 18.$

Hence, $a^2 + \dfrac{1}{a^2} = 18$.

(ii) By formula,

$\Rightarrow a^4 + \dfrac{1}{a^4} = \Big(a^2 + \dfrac{1}{a^2}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow a^4 + \dfrac{1}{a^4} = 18^2 - 2 \\[1em] = 324 - 2 \\[1em] = 322.$

Hence, $a^4 + \dfrac{1}{a^4} = 322$.

(iii) By formula,

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

Substituting values we get :

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

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