\(a,=\dfrac{1}{3}\cdot9\sqrt{3}-6\cdot\dfrac{2}{\sqrt{3}}+\dfrac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}}=3\sqrt{3}-4\sqrt{3}+\sqrt{3}+1=1\\ b,=\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}+\dfrac{14\left(3-\sqrt{2}\right)}{7}-\dfrac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{\sqrt{3}}\\ =\sqrt{6}+\sqrt{2}+6-2\sqrt{2}-\sqrt{6}+\sqrt{2}=6\)