Quick Algebra Identities

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

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

$\Rightarrow x^2+xy+y^2$

$\Rightarrow(x+y)^{2}-xy$

$\Rightarrow 10^{2}-1=\boxed{99}$

$\begin{aligned} x^2 + xy + y^2 & = (x+y)^2 - xy \\ & = \left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} + \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right)^2 - \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ & = \left(\frac{(\sqrt{3}-\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{3-2}\right)^2 - 1 \\ & = (2(3+2))^2 - 1 \\ & = 100 - 1 \\ & = \boxed{99} \end{aligned}$

Rationalize the denominator to get $x=5-2\sqrt {6}$ and $y=5+2\sqrt {6}$ . $x^{2}=49-20\sqrt{6}$ , $y^{2}=49+20\sqrt6$ and $xy=1$ . So the value of $x^2+y^2+xy=99$ .

Rationalizing the denominator :

$x = 5 - 2\sqrt{6}$

$y = 5 + 2\sqrt{6}$

$x^2 + y^2 + xy$ = $(x + y)^{2} - xy$ = $10^2 - 1$ = $99$

Rationalise x and y Add and subtract xy in x²+y²+xy Then solve it

here note that, y=1/x; so given expression becomes =(x^2)+1+(1/x^2); also note ,x+y=x+1/x =10; now consider [x+1/x]^2=(x^2)+2+(1/x^2)= given expression+1;

          so  100=given expression+1;
       hence, given expression=99: