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

This is also one good practice example that enlighten us we should keep an open-mindedness to solve for any given intriguing problem.

It'd be at first hard to foretell what would be the first step to start with this kind of problem. But, we know it won't be too much away to multiply the top and the bottom of the given fraction by some expression with the hope to simplify the nasty denominator as we have been given.

Personally I would ask my students to use the technique of substitution to make things more clearer and more manageable for them. But wait a minute, if we want the denominator be systematically expressed, then the last three terms should be expressed in terms of $\sqrt[4]{...}$, but looking at the given denominator the "awkward" terms $2\sqrt{5}$ and $\sqrt[4]{125}$ and it hasn't been "simplified" fully yet, so we get started by first rewriting the given fraction and we get:

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

$=\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2(\sqrt[4]{5})^2-(\sqrt[4]{5})^3}}$

Next, I would ask to substitute $x=\sqrt[4]{5}$ and then, here comes our first trial.

Our first trial would be to multiply both the top and the bottom by $1-\sqrt[4]{5}$, and there is no guarantee that this would work, let's see:

First trial with $\displaystyle \color{yellow}\bbox[5px,green]{1-\sqrt[4]{5}}$:

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

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1-x)}{\cdot (1-x)}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1-x)}{\cdot \sqrt{(1-x)^2}}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1-x)}{\cdot \sqrt{1-2x+x^2}}$

$=\dfrac{2(1-x)}{\sqrt{4-6x+4x^2-3x+6x^2-3x^3+2x^2-4x^3+2x^4-x^3+2x^4-x^5}}$

$=\dfrac{2(1-x)}{\sqrt{4-9x+12x^2-8x^3+4x^4-x^5}}(*)$

When we let $x=\sqrt[4]{5}$, we need to aware that 

$x^4=5$ and $x^5=5^\frac{5}{4}=5^{1+\frac{1}{4}}=5^{1}5^{\frac{1}{4}}=5x$

Therefore equation (*) becomes

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

$=\dfrac{2(1-x)}{\sqrt{4-9x+12x^2-8x^3+4x^4-x^5}}(*)$

$=\dfrac{2(1-x)}{\sqrt{4-9x+12x^2-8x^3+4(5)-5x}}$

$=\dfrac{2(1-x)}{\sqrt{24-14x+12x^2-8x^3}}$ (argh! Though this has not led to any solid effort yet, we don't give up, we could still try the other quantity $1+\sqrt[4]{5}$:

Second trial with $\displaystyle \color{yellow}\bbox[5px,purple]{1+\sqrt[4]{5}}$:

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

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1+x)}{\cdot (1+x)}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1+x)}{\cdot \sqrt{(1+x)^2}}$

$=\dfrac{2}{\sqrt{4-3x+2x^2-x^3}}\dfrac{\cdot (1+x)}{\cdot \sqrt{1+2x+x^2}}$

$=\dfrac{2(1-x)}{\sqrt{4+6x+4x^2-3x-6x^2-3x^3+2x^2+4x^3+2x^4-x^3-2x^4-x^5}}$

$=\dfrac{2(1-x)}{\sqrt{4+5x-x^5}}$ recall that $x^4=5$ and $x^5=5x$

$=\dfrac{2(1-x)}{\sqrt{4+5x-5x}}$

$=\dfrac{2(1-x)}{\sqrt{4}}$

$=\dfrac{\cancel{2}(1-x)}{\cancel{2}}$

$=1-x$

$=1-\sqrt[4]{5}$

And we can conclude by now that  $\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$ simplified to $1-\sqrt[4]{5}$.