How do I simplify this fraction with radicals?

[anemone]

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

 

[MarkFL]

My solution:

Spoiler

Let $$x=\sqrt[4]{5}$$

I would try to see if there is a solution for $a$ in:

$$\left(1+ax \right)^2\left(4-3x+2x^2-x^3 \right)=4$$

$$-a^2x^5+(2a^2-2a)x^4+(-3a^2+4a-1)x^3+(4a^2-6a+2)x^2+(8a-3)x+4=4$$

Now, we have:

$$x^5=5x$$ and $$x^4=5$$ hence we may write (after factoring):

$$(a-1)(1-3a)x^3+2(a-1)(2a-1)x^2+(a-1)(3-5a)x+10a(a-1)=0$$

Thus, we find $a=1$. And so we may write:

$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}$$

 

[anemone]

My solution:

Spoiler

Let $$x=\sqrt[4]{5}$$

I would try to see if there is a solution for $a$ in:

$$\left(1+ax \right)^2\left(4-3x+2x^2-x^3 \right)=4$$

$$-a^2x^5+(2a^2-2a)x^4+(-3a^2+4a-1)x^3+(4a^2-6a+2)x^2+(8a-3)x+4=4$$

Now, we have:

$$x^5=5x$$ and $$x^4=5$$ hence we may write (after factoring):

$$(a-1)(1-3a)x^3+2(a-1)(2a-1)x^2+(a-1)(3-5a)x+10a(a-1)=0$$

Thus, we find $a=1$. And so we may write:

$$\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}$$

Thanks for participating, MarkFL! You're very smart in checking if your first trial over the equality holds and if it does, what value of $a$ does it takes. Well done!