Algebra

Let's first rationalize the denominators of the equations.

$\frac{1}{\sqrt2+\sqrt3} = \frac{1}{\sqrt2+\sqrt3}\times1 = \frac{1}{\sqrt2+\sqrt3}\times\frac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3} = \frac{\sqrt2-\sqrt3}{(\sqrt{2}+\sqrt{3}) \times(\sqrt{2}-\sqrt{3})} = \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2}^2-\sqrt{3}^2} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = -(\sqrt{2} - \sqrt{3}) = \sqrt{3} - \sqrt{2}$

$\frac{1}{\sqrt{2}-\sqrt{3}} = \frac{1}{\sqrt{2}-\sqrt{3}} \times 1 = \frac{1}{\sqrt{2}-\sqrt{3}} \times \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} = \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}^2- \sqrt{3}^2} = \frac{\sqrt{2}+\sqrt{3}}{2-3} = -(\sqrt{2}+\sqrt{3}) = -\sqrt{2}-\sqrt{3}$

Now we can see that the origional equation is just (sqrt(3) - sqrt(2)) + (-sqrt(2) - sqrt(3)). This simplifies to -2sqrt(2)