If $\Big(x - \dfrac{1}{x}\Big) = 4$, find the values of :

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

(ii) $\Big(x^4 + \dfrac{1}{x^4}\Big)$.

(i) Given,

$\Big(x - \dfrac{1}{x}\Big) = 4$

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (4)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 - 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 16 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 16 + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 18$

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

(ii) Given,

$\Big(x - \dfrac{1}{x}\Big) = 4$

From part (i),

$x^2 + \dfrac{1}{x^2} = 18$

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (18)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 324 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 324 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 322.$

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