<s>[INST] Simplify $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$. 
Denest $\sqrt {\sqrt[3]{5}-\sqrt[3]{4}}$.

I have tried completing square by several method but all failed. Can anyone help me please? Thank you.
p.s. I'm a poor question-tagger.
 [/INST] There is a formula. I'm not too sure how to prove it, but I know that there is a formula where you can denest $$\sqrt{\sqrt[3]{\alpha}+\sqrt[3]{\beta}}$$Into$$\pm\frac {1}{\sqrt{f}}\left(-\frac {s^2\sqrt[3]{\alpha^2}}{2}+s\sqrt[3]{\alpha\beta}+\sqrt[3]{\beta^2}\right)$$ where $$f=\beta-s^3\alpha$$ and $s$ is a real number solution to $f(x)=x^4+4x^3+8\frac {\beta}{\alpha}x-4\frac {\beta}{\alpha}$
So in this case, $\alpha=5$ and $\beta=-4$. So $s=-2$ and $f=-4-(-2)^3\times 5=36$ Therefore, we have$$\pm\frac {1}{6}\left(-\frac {4\sqrt[3]{25}}{2}-2\sqrt[3]{-20}+\sqrt[3]{16}\right)=\pm\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$$
Discard the negative value to get $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$
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