Question

Simplify the following:
$\frac{6}{2\sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}$ = ?

Answer 1

We will first rationalize the denominators of the given expression and then solve it as follows:

$\frac{6}{2\sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}=(\frac{6}{2\sqrt{3}-\sqrt{6}}\times \frac{2\sqrt{3}+\sqrt{6}}{2\sqrt{3}+\sqrt{6}})+(\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})-(\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}\times \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}})=\left(\frac{6(2\sqrt{3}+\sqrt{6})}{(2\sqrt{3}-\sqrt{6})(2\sqrt{3}+\sqrt{6})}\right)+\left(\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\right)-\left(\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}\right)=\left(\frac{6(2\sqrt{3}+\sqrt{6})}{{\left(2\sqrt{3}\right)}^2-{\left(\sqrt{6}\right)}^2}\right)+\left(\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}\right)-\left(\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2}\right)\{\because (a-b)(a+b)=a^2-b^2\}$

$=\left(\frac{6(2\sqrt{3}+\sqrt{6})}{12-6}\right)+\left(\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{3-2}\right)-\left(\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{6-2}\right)=\left(\frac{6(2\sqrt{3}+\sqrt{6})}{6}\right)+\left(\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{1}\right)-\left(\frac{4\sqrt{3}(\sqrt{6}+\sqrt{2})}{4}\right)=(2\sqrt{3}+\sqrt{6})+\sqrt{6}(\sqrt{3}-\sqrt{2})-\sqrt{3}(\sqrt{6}+\sqrt{2})=(2\sqrt{3}+\sqrt{6})+(\sqrt{6\times 3}-\sqrt{6\times 2})-(\sqrt{3\times 6}+\sqrt{3\times 2})$

$=(2\sqrt{3}+\sqrt{6})+(\sqrt{18}-\sqrt{12})-(\sqrt{18}+\sqrt{6})=2\sqrt{3}+\sqrt{6}+\sqrt{18}-\sqrt{12}-\sqrt{18}-\sqrt{6}=2\sqrt{3}-\sqrt{12}=2\sqrt{3}-\sqrt{2\times 2\times 3}=2\sqrt{3}-2\sqrt{3}=0$

Hence, $\frac{6}{2\sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4\sqrt{3}}{\sqrt{6}-\sqrt{2}}=0$