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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, $x^2+xy+y^2=99$.