<s>[INST] Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let 
\begin{eqnarray}
x= \sqrt[3]{a} \\
      y= \sqrt[3]{b} \\
      z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that 
\begin{equation}
 x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z) 
\end{equation}
that turns out to be
\begin{equation}
(x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z)
\end{equation}
plug in $x+y+z$,
we will get 
\begin{equation}
\sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z)
\end{equation}
From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?
 [/INST] A systematic approach is to apply the denesting formula
$$\sqrt[3]{\sqrt[3]{A}-B} = \sqrt[3]{x_1}+ \sqrt[3]{x_2 }+  \sqrt[3]{x_3 } $$
where $x_1$, $x_2$ and $x_3$ are the roots of the cubic equation
$$x^3 + \frac{B+2C}3x^2 - \frac{(B-C)(2B+C)}{27}x+ \frac{(B-C)^3}{729}=0$$
with $C =\sqrt[3]{B^3-A}$. Thus, to denest $\sqrt[3]{\sqrt[3]{2}-1}$, solve
$$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}=0\implies
(x_1, x_2, x_3)= (\frac19,-\frac29,\frac49)$$
to obtain
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}&=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}
\end{align}
The general formula given above can be used as well to denest other nested cubic radicals, for instance
$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$
</s>