Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} \\ =\frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2-2\sqrt{6}}{3-2} \\ =5-2\sqrt{6} \end{gathered}$$

$$\begin{gathered} \mathrm{and} \\ b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ =\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{3+2+2\sqrt{6}}{3-2} \\ =5+2\sqrt{6} \end{gathered}$$

$$\begin{gathered} \mathrm{So},a^2+b^2-5ab={\left(5-2\sqrt{6}\right)}^2+{\left(5+2\sqrt{6}\right)}^2-5\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right) \\ =25+24-20\sqrt{6}+25+24+20\sqrt{6}-5\left[{\left(5\right)}^2-{\left(2\sqrt{6}\right)}^2\right] \\ =98-5\left(25-24\right) \\ =98-5 \\ =93 \end{gathered}$$