Question

Simplify the following:
$\frac{7\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{8}+\sqrt{2}}$.

Answer 1

We rationalise the denominators of $(\frac{7\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{2\sqrt{5}}{\sqrt{8}+\sqrt{2}})$ and then simplify as follows:

$\left(\frac{7\sqrt{3}}{\sqrt{6}-\sqrt{3}}\right)-\left(\frac{2\sqrt{5}}{\sqrt{8}+\sqrt{2}}\right)=(\frac{7\sqrt{3}}{\sqrt{6}-\sqrt{3}}\times \frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}})-(\frac{2\sqrt{5}}{\sqrt{8}+\sqrt{2}}\times \frac{\sqrt{8}-\sqrt{2}}{\sqrt{8}-\sqrt{2}})$

$=\left(\frac{7\sqrt{3}(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}\right)-\left(\frac{2\sqrt{5}(\sqrt{8}-\sqrt{2})}{(\sqrt{8}+\sqrt{2})(\sqrt{8}-\sqrt{2})}\right)$

$=\left(\frac{7\sqrt{18}+7\sqrt{9}}{(\sqrt{6}{)}^2-(\sqrt{3}{)}^2}\right)-\left(\frac{2\sqrt{40}-2\sqrt{10}}{(\sqrt{8}{)}^2-(\sqrt{2}{)}^2}\right)$

$=\frac{\left(7\sqrt{9\times 2}\right)+(7\times 3)}{6-3}-\frac{\left(2\sqrt{4\times 10}\right)-2\sqrt{10}}{8-2}=\frac{3(7\sqrt{2}+7)}{3}-\frac{2(2\sqrt{10}-\sqrt{10})}{6}$

$=7\sqrt{2}+7-\frac{\sqrt{10}}{3}$ .

Hence, $\left(\frac{7\sqrt{3}}{\sqrt{6}-\sqrt{3}}\right)-\left(\frac{2\sqrt{5}}{\sqrt{8}+\sqrt{2}}\right)=7\sqrt{2}+7-\frac{\sqrt{10}}{3}$ .