Given $x = \frac{1}{\sqrt{5}-\sqrt{3}}$ and $y = \frac{1}{\sqrt{5}+\sqrt{3}}$, we need to find the values of: (1) $x^2 + y^2$ (2) $x^3 + y^3$ (3) $\frac{y}{x} + \frac{x}{y}$ and match them with the options given.

1. Problem Description

Given $x = \frac{1}{\sqrt{5}-\sqrt{3}}$ and $y = \frac{1}{\sqrt{5}+\sqrt{3}}$, we need to find the values of:

(1) $x^2 + y^2$

(2) $x^3 + y^3$

(3) $\frac{y}{x} + \frac{x}{y}$

and match them with the options given.

2. Solution Steps

First, rationalize $x$ and $y$:

$x = \frac{1}{\sqrt{5}-\sqrt{3}} = \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})} = \frac{\sqrt{5}+\sqrt{3}}{5-3} = \frac{\sqrt{5}+\sqrt{3}}{2}$

$y = \frac{1}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{5}-\sqrt{3}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{\sqrt{5}-\sqrt{3}}{5-3} = \frac{\sqrt{5}-\sqrt{3}}{2}$

Now, calculate $x+y$ and $xy$:

$x+y = \frac{\sqrt{5}+\sqrt{3}}{2} + \frac{\sqrt{5}-\sqrt{3}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5}$

$xy = \frac{\sqrt{5}+\sqrt{3}}{2} \cdot \frac{\sqrt{5}-\sqrt{3}}{2} = \frac{5-3}{4} = \frac{2}{4} = \frac{1}{2}$

(1) $x^2 + y^2 = (x+y)^2 - 2xy = (\sqrt{5})^2 - 2(\frac{1}{2}) = 5 - 1 = 4$

The answer is ウ.

(2) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = (\sqrt{5})((\sqrt{5})^2 - 3(\frac{1}{2})) = (\sqrt{5})(5 - \frac{3}{2}) = \sqrt{5}(\frac{10-3}{2}) = \sqrt{5}(\frac{7}{2}) = \frac{7\sqrt{5}}{2}$

The answer is ク.

(3) $\frac{y}{x} + \frac{x}{y} = \frac{x^2 + y^2}{xy} = \frac{(x+y)^2 - 2xy}{xy} = \frac{(\sqrt{5})^2 - 2(\frac{1}{2})}{\frac{1}{2}} = \frac{5-1}{\frac{1}{2}} = \frac{4}{\frac{1}{2}} = 4 \cdot 2 = 8$

The answer is カ.

3. Final Answer

(1) $x^2 + y^2 = 4$ (ウ)

(2) $x^3 + y^3 = \frac{7\sqrt{5}}{2}$ (ク)

(3) $\frac{y}{x} + \frac{x}{y} = 8$ (カ)