Simplify $\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$ 
Simplify
  $$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$$

Found in a book with tag "Moscow 1982", the stated answer is $1+\sqrt[4]{5}$. Used all tricks that I know but without success. The answer appears to be correct, checked in Wolfram Alpha. 
Hints and answers welcomed. Sorry if this is a duplicate.
 A: Define $a=\sqrt[4]{5}$. Then $a^4 = 5$.
Consider the expression (under the radical in the denominator): $$
E = 4-3a+2a^2-a^3.\tag{1}
$$
What if we'll multiply it by $1+a$:
$$E(1+a)=(4-3a+2a^2-a^3)(1+a)\\=(4-3a+2a^2-a^3)+(4a-3a^2+2a^3-a^4)\\=4+a-a^2+a^3-a^4 \\= -1+a-a^2+a^3.\tag{2}
$$
Now what if we'll multiply it by $(1+a)$ again:
$$
E(1+a)^2 = (-1+a-a^2+a^3)(1+a)\\=(-1+a-a^2+a^3)+(-a+a^2-a^3+a^4)\\=
-1+a^4=4.\tag{3}$$
So,
$$
E=\dfrac{4}{(1+a)^2}.$$
Since $1+a>0$, we have $$\sqrt{E}=\dfrac{2}{1+a};$$
$$
\dfrac{2}{\sqrt{E}}=1+a=1+\sqrt[4]{5}.
$$
A: $$\begin{align}\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125})(\sqrt{1+\sqrt[4]{5}})}}\cdots \cdots(1)\\&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{5^3})+(4\cdot \sqrt[4]{5}-3\sqrt{5}+2\sqrt[4]{5^3}-5)}}\\&= \frac{2\cdot\sqrt{1+\sqrt[4]{5}}}{\sqrt{5^{1/4}-5^{1/2}+5^{3/4}-1}}\\&=\frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{(5^{1/4}-5^{1/2}+5^{3/4}-1)+5^{1/4}(5^{1/4}-5^{1/2}+5^{3/4}-1)}}\\&= \frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{4}}=1+\sqrt[4]{5}.\end{align} $$
at $(1)$ and 4th line multiplied the denominetor and numerator by $\sqrt{1+\sqrt[4]{5}}$.
