If $\Big(x + \dfrac{1}{x}\Big) = 5$, find the values of :

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

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

(i) Given,

$\Big(x + \dfrac{1}{x}\Big) = 5$

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + \Big(\dfrac{1}{x}\Big)^2 + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow (5)^2 = x^2 + \dfrac{1}{x^2} + 2 \times x \times \dfrac{1}{x} \\[1em] \Rightarrow 25 = x^2 + \dfrac{1}{x^2} + 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 25 - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = 23.$

Hence, $x^2 + \dfrac{1}{x^2} = 23$.

(ii) Given,

$\Big(x + \dfrac{1}{x}\Big) = 5$

From part (i),

$x^2 + \dfrac{1}{x^2} = 23$

Using identity,

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow (23)^2 = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] \Rightarrow 529 = x^4 + \dfrac{1}{x^4} + 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 529 - 2 \\[1em] \Rightarrow x^4 + \dfrac{1}{x^4} = 527 \\[1em]$

Hence, $x^4 + \dfrac{1}{x^4} = 527$.