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.
 A: See Johannes Blomer, How to denest Ramanujan's nested radicals, available here. 
A: 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 $\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac {1}{3}\left(-\sqrt[3]{25}+\sqrt[3]{20}+\sqrt[3]{2}\right)$
A: This example is discussed in one of my prior posts, based on a polynomial-time denesting algorithm of Blomer. Using standard Galois theory of radical (Kummer) extensions, it is not difficult to prove a  Denesting Structure Theorem, which implies that if a radical $\rm\; r^{1/d} \;$ denests in any radical extension $\rm\, F'$ of its base field $\rm\, F$, then a suitable multiple $\rm\; q b\: r \;$ of
the radicand $\rm\; r \;$ must already denest in the field $\rm\; F' \;$ defined by the
radicand. More precisely
Denesting Structure Theorem for Real Fields $\;\; \;$  Let $\rm\; F \;$ be a real field and
$\rm\; F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$
of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard
basis of $\rm\; F' \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F' \;$ and $\rm\; d \;$ is a positive integer such
that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is,
$\rm\; r^{1/d} \in F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for positive integers
$\rm\; t_i \;$ and positive $\rm\; a_i \in F,\:$ then there exists a nonzero $\rm\; q \in F \;$ and  $\rm\; b \in B \;$ with $\rm\; (q b r)^{1/d} \in F'.$
This implies that by multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$
and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can
normalize any denesting so that it denests in the field defined
by the radicand (then denesting reduces to solving for undetermined coefficients). For example
$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$
normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$
An example with nontrivial $\rm\:b$
$$  \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$
normalises to
$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$
Here $\rm\; F=\mathbb Q,\ F' = \mathbb Q(\sqrt 6),\ n=2,\ B = \{1,\sqrt 6\},\ d=2,\ q=1/3,\ b= \sqrt 6\:$.
The structure theorem also holds for complex fields except that
in this case one has to assume that $\rm\; F \;$ contains enough roots of
unity (which may be computationally expensive in practice, to
wit doubly-exponential complexity).
A: I will give you a hint. Seek denesting which looks like this :
$$ (\sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c})^2 = 9(\sqrt[3]{5}-\sqrt[3]{4}) $$
