Cube roots escape $ \sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{a} + \sqrt[3]{b} - \sqrt[3]{c}, $
where $ a, b $ and $ c $ are positive integers. What is the value of $ a+b+c $?
This question appeared in one of my exams
 A: You can check the following expression works by squaring the right hand side
$$
\sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25} \, .
$$
So $ a + b + c = 47 $.
As for finding it: It is initially a good idea to try working in the field extension $ \mathbb{Q}(\sqrt[3]{4}, \sqrt[3]{5}) = \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, which has a basis given by 
$$
\{ 1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{1\times5}, \sqrt[3]{2\times5}, \sqrt[3]{4\times5}, \sqrt[3]{1\times25}, \sqrt[3]{2\times25}, \sqrt[3]{4\times25} \} \, .
$$
Then you want to consider the equation
$$
  3\sqrt{\sqrt[3]{5} - \sqrt[3]{4}} = r + s \sqrt[3]{2} + t \sqrt[3]{4} + u \sqrt[3]{5} + v \sqrt[3]{10} + w \sqrt[3]{20} + x \sqrt[3]{25} + y \sqrt[3]{50} + z \sqrt[3]{100} \, ,
$$ with $ r, \ldots, z \in \mathbb{Q} $.  In particular you want $ r, \ldots, z \in \{ -1, 0, 1 \} $ with exactly three of them non-zero.  Square both sides, and compare the coefficient of $ \sqrt[3]{5} $ and $ \sqrt[3]{4} $ on both sides.  We get
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 2ru + 4tv + 4sw + 20yz \\
\sqrt[3]{4} &: \quad -9 = s^2 + 2rt + 10wx + 10vy + 10uz \, .
\end{align}
You've assumed that all the non-zero coefficients are $ \pm 1 $, so we must have $ x = \pm 1 $ and $ s = \pm 1 $, by parity considerations.  By assumption there are only 3 non-zero coefficients, $ s, x $ and one other.  So $ ru = tv = yz = 0 $ and $ rt = vy = uz = 0 $ and the equations reduce to
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 4sw \\
\sqrt[3]{4} &: \quad -9 = s^2 + 10wx \, .
\end{align}
We must have $ sw = 1 $ and $ wx = -1 $, so $ s = w = -x $.  This gives us two possibilities by taking $ x = 1 $ or $ x = - 1 $.  The form of the solution you are looking for has two $ +1 $ coefficients.  This must come from $ x = -1 $, and be
$$
  \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25}
$$
Now you need to go back and check this actually does work as a solution because we've made many assumptions in finding it.  In particular we've assumed that such a solution exists and lives in $ \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, rather than some larger field extension.  And we've only used two equations which come from comparing coefficients; do the rest also work?
A: Let
$ x^3=5$ 
$ y^3=4$ 
$ 9\sqrt{\strut x-y}$ 
$ =(x^3+y^3)\sqrt{\strut x-y}$ 
$ =(x^2-xy+y^2)\sqrt{\strut(x-y)(x+y)^2}$ 
$ =(x^2-xy+y^2)\sqrt{\strut x^3+x^2y-xy^2-y^3}$ 
Now we can substitute the value already
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut 5+\sqrt[3]{\strut 100}-\sqrt[3]{\strut80}-4}$ 
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut \sqrt[3]{\strut 10}^2-2\sqrt[3]{\strut10}+1}$ 
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})\sqrt{\strut (\sqrt[3]{\strut 10}-1)^2}$ 
$ =(\sqrt[3]{\strut 25}-\sqrt[3]{\strut 20}+\sqrt[3]{\strut 16})(\sqrt[3]{\strut 10}-1)$ 
$ =\sqrt[3]{\strut 250}-\sqrt[3]{\strut 200}+\sqrt[3]{\strut 160}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-\sqrt[3]{\strut 16}$ 
$ =5\sqrt[3]{\strut 2}-2\sqrt[3]{\strut 25}+2\sqrt[3]{\strut 20}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}-2\sqrt[3]{\strut 2}$ 
$ =3\sqrt[3]{\strut 2}-3\sqrt[3]{\strut 25}+3\sqrt[3]{\strut 20}$ 
Hence
$ 3\sqrt{\strut \sqrt[3]{\strut 5}-\sqrt[3]{\strut 4}}$ 
$ =\sqrt[3]{\strut 2}-\sqrt[3]{\strut 25}+\sqrt[3]{\strut 20}$ 
