If $x - \dfrac{1}{x} = \sqrt{5}$, find the values of

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

(ii) $x + \dfrac{1}{x}$

(iii) $x^3 + \dfrac{1}{x^3}$

(i) We know that,

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

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = (\sqrt{5})^2 + 2 \\[1em] = 5 + 2 \\[1em] = 7.$

Hence, $x^2 + \dfrac{1}{x^2}$ = 7.

(ii) We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \therefore x + \dfrac{1}{x} = \sqrt{x^2 + \dfrac{1}{x^2} + 2}$

Substituting values we get,

$x + \dfrac{1}{x} = \sqrt{7 + 2} \\[1em] = \sqrt{9} \\[1em] = \pm 3.$

Hence, $x + \dfrac{1}{x} = \pm 3.$

(iii) We know that,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \therefore x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big).$

When $x + \dfrac{1}{x} = 3$.

Substituting values we get,

$x^3 + \dfrac{1}{x^3} = 3^3 - 3 \times 3 \\[1em] = 27 - 9 \\[1em] = 18.$

When $x + \dfrac{1}{x} = -3$.

Substituting values we get,

$x^3 + \dfrac{1}{x^3} = (-3)^3 - 3 \times (-3) \\[1em] = -27 + 9 \\[1em] = -18.$

Hence, $x^3 + \dfrac{1}{x^3} = \pm 18$.