[b] If \( x=\sqrt{3}+\sqrt{2}, y=\frac{1}{\sqrt{3}+\sqrt{2}} \), find the value of: \( x^{2}-y^{2} \)

To find the value of \( x^2 - y^2 \), we can use the identity \( a^2 - b^2 = (a - b)(a + b) \). First, we calculate \( x^2 \) where \( x = \sqrt{3} + \sqrt{2} \): \[ x^2 = (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{3}\sqrt{2} = 5 + 2\sqrt{6}. \] Next, we find \( y \): \[ y = \frac{1}{\sqrt{3} + \sqrt{2}}. \] To simplify \( y \), we can multiply the numerator and the denominator by the conjugate of the denominator: \[ y = \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2}. \] Now, we calculate \( y^2 \): \[ y^2 = (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{3}\sqrt{2} = 5 - 2\sqrt{6}. \] Now we have \( x^2 \) and \( y^2 \): \[ x^2 = 5 + 2\sqrt{6}, \] \[ y^2 = 5 - 2\sqrt{6}. \] Finally, we compute \( x^2 - y^2 \): \[ x^2 - y^2 = (5 + 2\sqrt{6}) - (5 - 2\sqrt{6}) = 5 + 2\sqrt{6} - 5 + 2\sqrt{6} = 4\sqrt{6}. \] Thus, the answer is: \[ \boxed{4\sqrt{6}}. \]