find the value of

Solution:

According to question,

$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

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

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

$=\frac{5-2 \sqrt{6}}{1}$

$=5-2 \sqrt{6} \quad \ldots(1)$

$b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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

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

$=\frac{5+2 \sqrt{6}}{1}$

$=5+2 \sqrt{6} \quad \ldots(2)$

Now,

$a^{2}+b^{2}-5 a b$

$=(a-b)^{2}-3 a b$

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

$=(-4 \sqrt{6})^{2}-3(25-24)$

$=96-3$

$=93$

Hence, the value of $a^{2}+b^{2}-5 a b$ is 93 .