If $\Big(x - \dfrac{1}{x}\Big) = 8$, find the values of :

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

(ii) $\Big(x^2 - \dfrac{1}{x^2}\Big)$

(i) Given,

$\Big(x - \dfrac{1}{x}\Big) = 8$

We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - (8)^2 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - 64 = 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 64 + 4 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 68 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \sqrt{68} \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17} \\[1em]$

Hence, $\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}$.

(ii) Given,

$\Big(x - \dfrac{1}{x}\Big) = 8$

From (i),

$\Big(x + \dfrac{1}{x}\Big) = \pm 2\sqrt{17}$

Using identity,

$\Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \Big(x + \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = (\pm 2\sqrt{17})\times 8 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big) = \pm 16\sqrt{17}$

Hence, $x^2 - \dfrac{1}{x^2} = \pm 16\sqrt{17}$.