\( x=\frac{2}{\sqrt{10}-\sqrt{8}} ; y=\frac{2}{\sqrt{10}+2 \sqrt{2}} \) then \( (x-y)^{2} \)

To calculate \( (x-y)^2 \) where \( x=\frac{2}{\sqrt{10}-\sqrt{8}} \) and \( y=\frac{2}{\sqrt{10}+2 \sqrt{2}} \), we first simplify both \( x \) and \( y \). Starting with \( x \): \[ x = \frac{2}{\sqrt{10} - \sqrt{8}} \cdot \frac{\sqrt{10} + \sqrt{8}}{\sqrt{10} + \sqrt{8}} = \frac{2(\sqrt{10} + \sqrt{8})}{(\sqrt{10})^2 - (\sqrt{8})^2} = \frac{2(\sqrt{10} + \sqrt{8})}{10 - 8} \] \[ = \frac{2(\sqrt{10} + 2\sqrt{2})}{2} = \sqrt{10} + 2\sqrt{2} \] Now for \( y \): \[ y = \frac{2}{\sqrt{10} + 2\sqrt{2}} \cdot \frac{\sqrt{10} - 2\sqrt{2}}{\sqrt{10} - 2\sqrt{2}} = \frac{2(\sqrt{10} - 2\sqrt{2})}{(\sqrt{10})^2 - (2\sqrt{2})^2} = \frac{2(\sqrt{10} - 2\sqrt{2})}{10 - 8} \] \[ = \frac{2(\sqrt{10} - 2\sqrt{2})}{2} = \sqrt{10} - 2\sqrt{2} \] Now, calculating \( x - y \): \[ x - y = (\sqrt{10} + 2\sqrt{2}) - (\sqrt{10} - 2\sqrt{2}) = 4\sqrt{2} \] Finally, we find \( (x - y)^2 \): \[ (x - y)^2 = (4\sqrt{2})^2 = 16 \cdot 2 = 32 \] The answer is \( \boxed{32} \).