If $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=x,\ \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=y$, find the value $x^{2}+y^{2}+x y$.

Given: $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=x,\ \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=y$.

To do: To find the value $x^{2}+y^{2}+x y$.

Solution:

As given $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=x,\ \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=y$

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

$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$    [On multiplying both numerators and denominators by $\sqrt{3}+\sqrt{2}$]

$=\frac{( \sqrt{3}+\sqrt{2})^2}{( ( \sqrt{3})^2-( \sqrt{2})^2)}$

$=\frac{( \sqrt{3})^2+2\sqrt{3}\sqrt{2}+( \sqrt{2})^2}{3-2}$

$=\frac{3+2\sqrt{6}+2}{1}$

$=5+2\sqrt{6}$

$\Rightarrow x^2=( 5+2\sqrt{6})^2$

$\Rightarrow x^2=5^2+2\times5\times2\sqrt{6}+( 2\sqrt{6})^2$

$\Rightarrow x^2=25+20\sqrt{6}+24$

$\Rightarrow x^2=49+20\sqrt{6}$

Similarly, $y^2=49-20\sqrt{6}$

Therefore, $x^2+xy+y^2=49+\sqrt{6}+( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}})( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}})+49-\sqrt{6}$

$=98+1$

$=99$

Thus,  $x^2+xy+y^2=99$.