If x = $\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \text{ and } y = \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}$; find :

(i) x²

(ii) y²

(iii) xy

(iv) x² + y² + xy

(i) Substituting value of x, we get :

$\Rightarrow x^2 = \Big(\dfrac{\sqrt{5} - 2}{\sqrt{5} + 2}\Big)^2 \\[1em] = \Big(\dfrac{\sqrt{5}^2 + 2^2 - 2 \times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 + 2 \times \sqrt{5} \times 2}\Big) \\[1em] = \dfrac{5 + 4 - 4\sqrt{5}}{5 + 4 + 4\sqrt{5}} \\[1em] = \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}}.$

Rationalizing,

$= \dfrac{9 - 4\sqrt{5}}{9 + 4\sqrt{5}} \times \dfrac{9 - 4\sqrt{5}}{9 - 4\sqrt{5}} \\[1em] = \dfrac{(9 - 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 - 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 - 72\sqrt{5}}{1} \\[1em] = 161 - 72\sqrt{5}.$

Hence, x² = $161 - 72\sqrt{5}.$

(ii) Substituting value of y, we get :

$\Rightarrow y^2 = \Big(\dfrac{\sqrt{5} + 2}{\sqrt{5} - 2}\Big)^2 \\[1em] = \dfrac{(\sqrt{5})^2 + 2^2 + 2\times \sqrt{5} \times 2}{(\sqrt{5})^2 + 2^2 - 2\times \sqrt{5} \times 2} \\[1em] = \dfrac{5 + 4 + 4\sqrt{5}}{5 + 4 - 4\sqrt{5}} \\[1em] = \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}}$

Rationalizing,

$= \dfrac{9 + 4\sqrt{5}}{9 - 4\sqrt{5}} \times \dfrac{9 + 4\sqrt{5}}{9 + 4\sqrt{5}} \\[1em] = \dfrac{(9 + 4\sqrt{5})^2}{9^2 - (4\sqrt{5})^2} \\[1em] = \dfrac{9^2 + (4\sqrt{5})^2 + 2 \times 9 \times 4\sqrt{5}}{81 - 80} \\[1em] = \dfrac{81 + 80 + 72\sqrt{5}}{1} \\[1em] = 161 + 72\sqrt{5}.$

Hence, y² = $161 + 72\sqrt{5}.$

(iii) Substituting values of x and y, we get :

$\Rightarrow xy = \dfrac{\sqrt{5} - 2}{\sqrt{5} + 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} - 2} \\[1em] = 1.$

Hence, xy = 1.

(iv) Substituting value of x², y² and xy, we get :

$\Rightarrow x^2 + y^2 + xy = 161 - 72\sqrt{5} + 161 + 72\sqrt{5} + 1 \\[1em] = 323.$

Hence, x² + y² + xy = 323.