Question

If $a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\mathrm{and}b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$, find the value of a² + b² – 5ab.

Answer 1

According to question,
$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\mathrm{and}b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2-2\sqrt{6}}{3-2} \\ =\frac{5-2\sqrt{6}}{1} \\ =5-2\sqrt{6}....\left(1\right) \\ b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\left(\sqrt{3}\right)\left(\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2+2\sqrt{6}}{3-2} \\ =\frac{5+2\sqrt{6}}{1} \\ =5+2\sqrt{6}....\left(2\right) \end{gathered}$$
Now,
$$\begin{gathered} a^2+b^2-5ab \\ ={\left(a-b\right)}^2-3ab \\ ={\left\{\left(5-2\sqrt{6}\right)-\left(5+2\sqrt{6}\right)\right\}}^2-3\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right) \\ ={\left(-4\sqrt{6}\right)}^2-3\left(25-24\right) \\ =96-3 \\ =93 \end{gathered}$$

Hence, the value of a² + b² – 5ab is 93.