<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] An observation:
\begin{align*}
(\sqrt[3]{4}-\sqrt[3]{2}+1)^3&=\frac{(\sqrt[3]{8}+1)^3}{(\sqrt[3]{2}+1)^3}\\
&=\frac{27}{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}\\
&=\frac{9}{\sqrt[3]{4}+\sqrt[3]{2}+1}\\
&=\frac{9(\sqrt[3]{2}-1)}{\sqrt[3]{8}-1}\\
&=9(\sqrt[3]{2}-1)\\
\left(\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{-2}{9}}+\sqrt[3]{\frac{1}{9}}\right)^3&=\sqrt[3]{2}-1
\end{align*}
</s>