Question

Rationalize the denominator and simply
$\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

Answer 1

$\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

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

$\frac{3\sqrt{2}(\sqrt{3}-\sqrt{6})}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{6}\right)}^2}-\frac{4\sqrt{3}(\sqrt{6}-\sqrt{2})}{{\left(\sqrt{6}\right)}^2-{\left(\sqrt{2}\right)}^2}+\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2}$

$=\frac{3\sqrt{2\times 3}-3\sqrt{2\times 6}}{3-6}-\frac{4\sqrt{3\times 6}-4\sqrt{3\times 2}}{6-2}+\frac{\sqrt{6\times 2}-\sqrt{6\times 3}}{2-3}$

$=\frac{3\sqrt{6}-3\sqrt{12}}{-3}-\frac{4\sqrt{18}-4\sqrt{6}}{4}+\frac{\sqrt{12}-\sqrt{18}}{-1}$

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

$=\frac{24\sqrt{3}-12\sqrt{6}+12\sqrt{6}-36\sqrt{2}+36\sqrt{2}-24\sqrt{3}}{12}$

$=0$

Hence the correct answer is $0$.