<s>[INST] 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.
 [/INST] $$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2\sqrt{6}}$$
Cancelling those $2\sqrt{6}$ you would end up with required result...
Can you see it at least now?
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