Question
Simplify: $\frac{2\sqrt{6}}{{\sqrt2}+\sqrt{3}}+\frac{6\sqrt{2}}{{\sqrt6}+\sqrt{3}}-\frac{8\sqrt{3}}{{\sqrt6}+\sqrt{2}}$

Answer

$\frac{2\sqrt{6}}{{\sqrt2}+\sqrt{3}}+\frac{6\sqrt{2}}{{\sqrt6}+\sqrt{3}}-\frac{8\sqrt{3}}{{\sqrt6}+\sqrt{2}}$
$=\frac{2\sqrt{6}}{{\sqrt2}+\sqrt{3}}\times\frac{{\sqrt2}-\sqrt{3}}{{\sqrt2}-\sqrt{3}}+\frac{6\sqrt{2}}{{\sqrt6}+\sqrt{3}}\times\frac{{\sqrt6}-\sqrt{3}}{{\sqrt6}-\sqrt{3}}-\frac{8\sqrt{3}}{{\sqrt6}+\sqrt{2}}\times\frac{{\sqrt6}-\sqrt{2}}{{\sqrt6}-\sqrt{2}}$
$=\frac{2\sqrt{6}\times\sqrt{3}-2\sqrt{6}\times\sqrt{2}}{\big(\sqrt{3}\big)^2-\big(\sqrt{2}\big)^2}+\frac{6\sqrt{2}\times\sqrt{6}-6\sqrt{2}\times\sqrt{3}}{\big(\sqrt{6}\big)^2-\big(\sqrt{3}\big)^2}-\frac{8\sqrt{3}\times\sqrt{6}-8\sqrt{3}\times\sqrt{2}}{\big(\sqrt{6}\big)^2-\big(\sqrt{2}\big)^2}$
$=\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2}$
$=2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4}$
$=2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6}$
$=0$

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