<s>[INST] 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.
 [/INST] $$\begin{align}\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125})(\sqrt{1+\sqrt[4]{5}})}}\cdots \cdots(1)\\&= \frac{2\cdot \sqrt{1+\sqrt[4]{5}}}{\sqrt{(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{5^3})+(4\cdot \sqrt[4]{5}-3\sqrt{5}+2\sqrt[4]{5^3}-5)}}\\&= \frac{2\cdot\sqrt{1+\sqrt[4]{5}}}{\sqrt{5^{1/4}-5^{1/2}+5^{3/4}-1}}\\&=\frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{(5^{1/4}-5^{1/2}+5^{3/4}-1)+5^{1/4}(5^{1/4}-5^{1/2}+5^{3/4}-1)}}\\&= \frac{2\cdot (1+\sqrt[4]{5})}{\sqrt{4}}=1+\sqrt[4]{5}.\end{align} $$
at $(1)$ and 4th line multiplied the denominetor and numerator by $\sqrt{1+\sqrt[4]{5}}$.
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