<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] I try this:
$y=\sqrt[3]{2}$,$y^3=2$, $x= \sqrt[3]{\sqrt[3]{2}-1}$
Therefore, $x^3=y-1$
$y^3-1=(y-1)(y^2+y+1)=1$
Because $y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$
and $y^3+1=3=(y+1)(y^2-y+1)$ 
Hence 
$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}=\frac{1}{9}(y^2-y+1)$
Therefore, based on the definition of $y$, we can obtain 
$a=\frac{4}{9}$,$b=-\frac{2}{9}$, and $c=\frac{1}{9}$
</s>