Given: \(y = \cfrac{1}{\sqrt{3} + \sqrt{2}}\) \(= \cfrac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}\) \(= \cfrac{\sqrt{3} - \sqrt{2}}{3 - 2}\) \(= \sqrt{3} - \sqrt{2}\) Therefore: \(x + y = \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} = 2\sqrt{3}\) and \(x - y = \sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2} = 2\sqrt{2}\) So, \((x + y)^2 + (x - y)^2 = (2\sqrt{3})^2 + (2\sqrt{2})^2\) \(= 12 + 8 = 20\) \(12 - 8 = 4\), Answer