Simplify: $\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$ I am doing a pretty hard problem:
$$\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
So it is a pretty long and complicated problem. I got stuck though. My idea was to turn $\sqrt[4]{27} =\sqrt[4]{3}\sqrt{3}$ and since I cant make the second part easier with Langranges formula (it doesn't apply to this) I made it $\sqrt[4]{3}-1$.
I seemed happy that I was getting somewhere and I thought that I had it but later on I just got stuck primarily by the 1's that I don't know what to do with. 
 A: Let $$x=\frac{\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}}$$
The square of numerator is $$2\sqrt[4]{27} -2\sqrt{\sqrt{27}-(\sqrt{3}-1)}=2\sqrt[4]{27}-2\sqrt{2\sqrt{3}+1}$$
The square of denominator is $\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}$. 
Hence $x^2 = 2$. Also note that $x>0$.
A: \begin{align}
&=\frac{\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}-\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\bigg)\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber \\
&=\frac{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}-\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber\\
&=\frac{2\sqrt{\sqrt{3}-1}}{\sqrt{\sqrt[4]{27}-\sqrt{2\sqrt{3}+1}}\bigg(\sqrt{\sqrt[4]{27}+\sqrt{\sqrt{3}-1}}+\sqrt{\sqrt[4]{27}-\sqrt{\sqrt{3}-1}}\big)}\nonumber
\end{align}
Can you take it from here?
