Question

Show that : $\frac{3\sqrt{2}}{3+\sqrt{6}}+\frac{6}{2\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{2}+\sqrt{6}}=0$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{3\sqrt{2}}{3+\sqrt{6}}+\frac{6}{2\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{2}+\sqrt{6}} \\ =\frac{3\sqrt{2}}{3+\sqrt{6}}\times \frac{3-\sqrt{6}}{3-\sqrt{6}}+\frac{6}{2\sqrt{3}+\sqrt{6}}\times \frac{2\sqrt{3}-\sqrt{6}}{2\sqrt{3}-\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{2}+\sqrt{6}}\times \frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}} \\ =\frac{9\sqrt{2}-6\sqrt{3}}{{\left(3\right)}^2-{\left(\sqrt{6}\right)}^2}+\frac{12\sqrt{3}-6\sqrt{6}}{{\left(2\sqrt{3}\right)}^2-{\left(\sqrt{6}\right)}^2}-\frac{4\sqrt{6}-12\sqrt{2}}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{6}\right)}^2} \\ =\frac{9\sqrt{2}-6\sqrt{3}}{9-6}+\frac{12\sqrt{3}-6\sqrt{6}}{12-6}-\frac{4\sqrt{6}-12\sqrt{2}}{2-6} \\ =\frac{9\sqrt{2}-6\sqrt{3}}{3}+\frac{12\sqrt{3}-6\sqrt{6}}{6}+\frac{4\sqrt{6}-12\sqrt{2}}{4} \\ =\frac{36\sqrt{2}-24\sqrt{3}+24\sqrt{3}-12\sqrt{6}+12\sqrt{6}-36\sqrt{2}}{12} \\ =0 \\ \mathrm{RHS}=0=\text{LHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$