If x = 3 + 1 3 - 1 and y = 3 - 1 3 + 1, then the value of ( x^2 + y^2 ) is?

Answer: Option C

$$\begin{array}{rl}& x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}\\ & =\frac{{\left(\sqrt{3}+1\right)}^2}{3-1}\\ & =\frac{3+1+2\sqrt{3}}{2}\\ & =2+\sqrt{3}\\ & y=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\\ & =\frac{{\left(\sqrt{3}-1\right)}^2}{3-1}\\ & =\frac{3+1-2\sqrt{3}}{2}\\ & =2-\sqrt{3}\\ & \therefore x^2+y^2={\left(2+\sqrt{3}\right)}^2+{\left(2-\sqrt{3}\right)}^2\\ & =2\left(4+3\right)\\ & =14\end{array}$$