Question

\(\dfrac{3}{3-\sqrt{6}}+\dfrac{3\sqrt{3}+2\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)

Answer 1

\(=\dfrac{3\left(3+\sqrt{6}\right)}{3}+\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{6}+2\right)}{\sqrt{3}+\sqrt{2}}\\ =\sqrt{3}+\sqrt{6}+5-\sqrt{6}=5+\sqrt{3}\)

Answer 2

\(\dfrac{3}{3-\sqrt{6}}+\dfrac{3\sqrt{3}+2\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\dfrac{3}{3-\sqrt{6}}+\dfrac{\left(\sqrt{3}\right)^3+\left(\sqrt{2}\right)^3}{\sqrt{3}+\sqrt{2}}=\dfrac{3\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}+\dfrac{\left(\sqrt{3}+\sqrt{2}\right)\left(3-\sqrt{6}+2\right)}{\sqrt{3}+\sqrt{2}}\)

\(=\dfrac{3\left(3+\sqrt{6}\right)}{9-6}+5-\sqrt{6}=3+\sqrt{6}+5-\sqrt{6}=8\)