Rationalization of $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ 
Question:
$$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ equals:

My approach:
I tried to rationalize the denominator by multiplying it by $\frac{\sqrt{2}-\sqrt{3}-\sqrt{5}}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$. And got the result to be (after a long calculation):
$$\frac{\sqrt{24}+\sqrt{40}-\sqrt{16}}{\sqrt{12}+\sqrt{5}}$$
which is totally not in accordance with the answer, $\sqrt{2}+\sqrt{3}-\sqrt{5}$.
Can someone please explain this/give hints to me.

Your "long calculation" was obviously wrong, since the denominator should be
$$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$
