An algebra problem by Abhiram Rao

$\begin{aligned} x+y & = \frac {\sqrt 3 + \sqrt 2}{\sqrt 3 - \sqrt 2} + \frac {\sqrt 3 - \sqrt 2}{\sqrt 3 + \sqrt 2} \\ & = \frac {(\sqrt 3 + \sqrt 2)^2 + (\sqrt 3 - \sqrt 2)^2}{(\sqrt 3 + \sqrt 2)(\sqrt 3 - \sqrt 2)} \\ & = \frac {5+2\sqrt 6 + 5-2\sqrt 6}{1} \\ & = 10 \\ (x+y)^2 & = x^2 + y^2 + 2xy \\ 10^2 & = x^2 + y^2 + 2 \times \frac {\sqrt 3 + \sqrt 2}{\sqrt 3 - \sqrt 2} \times \frac {\sqrt 3 - \sqrt 2}{\sqrt 3 + \sqrt 2} \\ 100 & = x^2 + y^2 + 2 \\ \implies x^2 + y^2 & = 100 -2 = \boxed{98} \end{aligned}$

$x = \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{(\sqrt{3}+\sqrt{2})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\\ =\dfrac{(\sqrt{3}+\sqrt{2})^2}{3-2}\\ =5+2\sqrt{6}$

$y=\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\\ =\dfrac{(\sqrt{3}-\sqrt{2})(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\ =\dfrac{(\sqrt{3}-\sqrt{2})^2}{3-2}\\ =5-2\sqrt{6}$

$x^2+y^2\\ =(5+2\sqrt{6})^2+(5-2\sqrt{6})^2\\ =25+4(6)\color{#D61F06}{+20\sqrt{6}}+25+4(6)\color{#D61F06}{-20\sqrt{6}}\\ =25+24+25+24\\ =\boxed{98}$