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.

Hint $\ (\sqrt a\!+\!\sqrt b\, + \sqrt c)(\sqrt a\! +\!\sqrt b\, - \sqrt c)\, =\, a\!+\!b\!-\!c+2\sqrt{ab}\ \ (=\, 2\sqrt{ab}\ \ {\rm if}\ \ a+b=c)$
