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.
 A: Your "long calculation" was obviously wrong, since the denominator should be
$$(\sqrt 2 + \sqrt 3 + \sqrt 5)(\sqrt 2 - (\sqrt 3 + \sqrt 5))=\\=\sqrt2^2 - (\sqrt 3 + \sqrt 5)^2 = 2 - (3 + 5 + 2\sqrt{15}) = -6-2\sqrt{15}$$
A: $$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3})^2-5}=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{5+2\sqrt{6}-5}$$
$$=\frac{2\sqrt{6}\cdot(\sqrt{2}+\sqrt{3}-\sqrt{5})}{2\sqrt{6}}$$
Cancelling those $2\sqrt{6}$ you would end up with required result...
Can you see it at least now?
A: What I would do is multiply by the first term plus the conjugate of the last two terms. I have coloured the important parts of the following expression to make it easier to understand.
$$\frac{2\sqrt{6}}{\color{green}{\sqrt{2}+\sqrt{3}+}\color{red}{\sqrt{5}}}\cdot\frac{\color{green}{\sqrt{2}+\sqrt{3}-}\color{red}{\sqrt{5}}}{\color{green}{\sqrt{2}+\sqrt{3}-}\color{red}{\sqrt{5}}}$$
$$\frac{2\sqrt{6}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}-\sqrt{5})}$$
Why do I do this, you ask? Remember the difference of squares formula:
$$a^2-b^2=(a+b)(a-b)$$
I am actually letting $a=\color{green}{\sqrt{2}+\sqrt{3}}$ and $b=\color{red}{\sqrt{5}}$. Therefore, our fraction can be rewritten as:
$$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\color{green}{\sqrt{2}+\sqrt{3}})^2-(\color{red}{\sqrt{5}})^2}$$
$$=\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2+2\sqrt{6}+3-5}$$
Oh. How nice. The integers in the denominator cancel out!
$$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}$$
Multiply by $\dfrac{2\sqrt{6}}{2\sqrt{6}}$
$$\frac{4\sqrt{3}+6\sqrt{2}-2\sqrt{30}}{2\sqrt{6}}\cdot\frac{2\sqrt{6}}{2\sqrt{6}}$$
$$=\frac{8\sqrt{3}\sqrt{6}+12\sqrt{2}\sqrt{6}-4\sqrt{30}\sqrt{6}}{(2\sqrt{6})^2}$$
$$=\frac{\color{red}{24}\sqrt{2}+\color{red}{24}\sqrt{3}-\color{red}{24}\sqrt{5}}{\color{red}{24}}$$
$$=\frac{\color{red}{24}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{24}}$$
Cancel $24$ out in the numerator and denominator and you get:
$$\sqrt{2}+\sqrt{3}-\sqrt{5}$$
$$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$

There is actually a much shorter way. Let's go back to the fraction
$$\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$
$$=\frac{2\sqrt{6}\sqrt{2}+2\sqrt{6}\sqrt{3}-2\sqrt{6}\sqrt{5}}{2+2\sqrt{2}\sqrt{3}+3-5}$$
$$=\frac{\color{red}{2\sqrt{6}}\sqrt{2}+\color{red}{2\sqrt{6}}\sqrt{3}-\color{red}{2\sqrt{6}}\sqrt{5}}{\color{red}{2\sqrt{6}}}$$
Do you see that we can factor out $2\sqrt{6}$ in the numerator?
$$\frac{\color{red}{2\sqrt{6}}(\sqrt{2}+\sqrt{3}-\sqrt{5})}{\color{red}{2\sqrt{6}}}$$
Cancel $2\sqrt{6}$ in the numerator and the denominator out, and you get:
$$\sqrt{2}+\sqrt{3}-\sqrt{5}$$
$$\displaystyle \color{green}{\boxed{\therefore \dfrac{2\sqrt{6}}{\sqrt{2}+\sqrt3+\sqrt5}=\sqrt{2}+\sqrt{3}-\sqrt{5}}}$$

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