If m = $\dfrac{1}{3 - 2\sqrt{2}} \text{ and } n = \dfrac{1}{3 + 2\sqrt{2}}$, find :

(i) m²

(ii) n²

(iii) mn

(i) Substituting value of m, we get :

$\Rightarrow m^2 = \Big(\dfrac{1}{3 - 2\sqrt{2}}\Big)^2 \\[1em] = \dfrac{1^2}{(3 - 2\sqrt{2})^2} \\[1em] = \dfrac{1}{3^2 + (2\sqrt{2})^2 - 2 \times 3 \times 2\sqrt{2}} \\[1em] = \dfrac{1}{9 + 8 - 12\sqrt{2}} \\[1em] = \dfrac{1}{17 - 12\sqrt{2}}$

Rationalizing,

$= \dfrac{1}{17 - 12\sqrt{2}} \times \dfrac{17 + 12\sqrt{2}}{17 + 12\sqrt{2}} \\[1em] = \dfrac{17 + 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \dfrac{17 + 12\sqrt{2}}{289 - 288} \\[1em] = \dfrac{17 + 12\sqrt{2}}{1} \\[1em] = 17 + 12\sqrt{2}.$

Hence, m² = $17 + 12\sqrt{2}$.

(ii) Substituting value of n, we get :

$\Rightarrow n^2 = \Big(\dfrac{1}{3 + 2\sqrt{2}}\Big)^2 \\[1em] = \dfrac{1^2}{(3 + 2\sqrt{2})^2} \\[1em] = \dfrac{1}{3^2 + (2\sqrt{2})^2 + 2 \times 3 \times 2\sqrt{2}} \\[1em] = \dfrac{1}{9 + 8 + 12\sqrt{2}} \\[1em] = \dfrac{1}{17 + 12\sqrt{2}}$

Rationalizing,

$= \dfrac{1}{17 + 12\sqrt{2}} \times \dfrac{17 - 12\sqrt{2}}{17 - 12\sqrt{2}} \\[1em] = \dfrac{17 - 12\sqrt{2}}{17^2 - (12\sqrt{2})^2} \\[1em] = \dfrac{17 - 12\sqrt{2}}{289 - 288} \\[1em] = \dfrac{17 - 12\sqrt{2}}{1} \\[1em] = 17 - 12\sqrt{2}.$

Hence, n² = $17 - 12\sqrt{2}$.

(iii) Substituting value of m and n, we get :

$\Rightarrow mn = \dfrac{1}{3 - 2\sqrt{2}} \times \dfrac{1}{3 + 2\sqrt{2}} \\[1em] = \dfrac{1}{3^2 + 3 \times 2\sqrt{2} - 2\sqrt{2} \times 3 - (2\sqrt{2})^2} \\[1em] = \dfrac{1}{9 + 6\sqrt{2} - 6\sqrt{2} - 8} \\[1em] = \dfrac{1}{9 - 8} \\[1em] = \dfrac{1}{1} \\[1em] = 1.$

Hence, mn = 1.