trig and solve

$\frac{2-\sqrt{3}}{\sqrt{6} - (\sqrt{2}+1)} \cdot \frac{\sqrt{6} + (\sqrt{2}+1)}{\sqrt{6} + (\sqrt{2}+1)}$

$\frac{2\sqrt{6} + 2\sqrt{2} + 2 - 3\sqrt{2} - \sqrt{6} - \sqrt{3}}{6 - (2 + 2\sqrt{2} + 1)}$

$\frac{\sqrt{6} - \sqrt{2} - \sqrt{3} + 2}{3 - 2\sqrt{2}} \cdot \frac{3 + 2\sqrt{2}}{3 + 2\sqrt{2}}$

$\frac{3\sqrt{6} - 3\sqrt{2} - 3\sqrt{3} + 6 + 4\sqrt{3} - 4 - 2\sqrt{6} + 4\sqrt{2}}{9-8}$

$\sqrt{6} + \sqrt{2} + \sqrt{3} + 2$