If $x - 2 = \dfrac{1}{3x}$, find the values of :

(i) $\Big(x^2 + \dfrac{1}{9x^2}\Big)$

(ii) $\Big(x^4 + \dfrac{1}{81x^4}\Big)$.

(i) Given,

$\Rightarrow x - 2 = \dfrac{1}{3x} \\[1em] \Rightarrow x - \dfrac{1}{3x} = 2$

We know that,

$\Rightarrow \Big(x - \dfrac{1}{3x}\Big)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow (2)^2 = x^2 + \Big(\dfrac{1}{3x}\Big)^2 - 2 \times x \times \dfrac{1}{3x} \\[1em] \Rightarrow 4 = x^2 + \dfrac{1}{9x^2} - \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = 4 + \dfrac{2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{12 + 2}{3} \\[1em] \Rightarrow x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$

Hence, $x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$.

(ii) From part (i),

$x^2 + \dfrac{1}{9x^2} = \dfrac{14}{3}$

Using identity,

$\Rightarrow \Big(x^2 + \dfrac{1}{9x^2}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \Big(\dfrac{14}{3}\Big)^2 = (x^2)^2 + \Big(\dfrac{1}{9x^2}\Big)^2 + 2 \times x^2 \times \dfrac{1}{9x^2} \\[1em] \Rightarrow \dfrac{196}{9} = x^4 + \dfrac{1}{81x^4} + \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196}{9} - \dfrac{2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{196 - 2}{9} \\[1em] \Rightarrow x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}$

Hence, $x^4 + \dfrac{1}{81x^4} = \dfrac{194}{9}$.