If x = $2\sqrt{3} + 2\sqrt{2}$, find :

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

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

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

(i) Substituting value of x, we get :

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{2\sqrt{3} + 2\sqrt{2}}$

Rationalizing,

$= \dfrac{1}{2\sqrt{3} + 2\sqrt{2}} \times \dfrac{2\sqrt{3} - 2\sqrt{2}}{2\sqrt{3} - 2\sqrt{2}} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{(2\sqrt{3})^2 - (2\sqrt{2})^2} \\[1em] = \dfrac{2\sqrt{3} - 2\sqrt{2}}{12 - 8} \\[1em] = \dfrac{2(\sqrt{3} - \sqrt{2})}{4} \\[1em] = \dfrac{\sqrt{3} - \sqrt{2}}{2}.$

Hence, $\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{2}$.

(ii) Substituting values of x and $\dfrac{1}{x}$ we get :

$\Rightarrow x + \dfrac{1}{x} = 2\sqrt{3} + 2\sqrt{2} + \dfrac{\sqrt{3} - \sqrt{2}}{2} \\[1em] = \dfrac{4\sqrt{3} + 4\sqrt{2} + \sqrt{3} -\sqrt{2}}{2} \\[1em] = \dfrac{5\sqrt{3} + 3\sqrt{2}}{2}.$

Hence, $x + \dfrac{1}{x} = \Big(\dfrac{5\sqrt{3} + 3\sqrt{2}}{2}\Big)$.

(iii) Substituting value of $\Big(x + \dfrac{1}{x}\Big)$, we get :

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = \Big(\dfrac{5\sqrt{3} + 3\sqrt{2}}{2}\Big)^2 \\[1em] = \dfrac{(5\sqrt{3})^2 + (3\sqrt{2})^2 + 2 \times 5\sqrt{3} \times 3\sqrt{2}}{2^2} \\[1em] = \dfrac{75 + 18 + 30\sqrt{6}}{4} \\[1em] = \dfrac{93 + 30\sqrt{6}}{4}.$

Hence, $\Big(x + \dfrac{1}{x}\Big)^2 = \dfrac{93 + 30\sqrt{6}}{4}$.