\(=2\cdot\dfrac{4}{\sqrt{3}}-3\cdot\dfrac{1}{3\sqrt{3}}-6\cdot\dfrac{2}{3\sqrt{5}}\\ =\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{4}{\sqrt{5}}=\dfrac{7}{\sqrt{3}}-\dfrac{4\sqrt{5}}{5}\\ =\dfrac{7\sqrt{3}}{3}-\dfrac{4\sqrt{5}}{5}=\dfrac{35\sqrt{3}-12\sqrt{5}}{15}\)