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?

Here is an alternative to denesting $(2^{1/3}-1)^{1/3}$. First, set $x^3=2$ so that$$x^3-1=1$$Factoring the left-hand side and isolating $x-1$ gives$$x-1=\frac 1{1+x+x^2}=\frac 3{3+x+3x^2}=\frac 3{(1+x)^3}$$Multiply both sides by $9$ to complete the cube$$9(x-1)=\left(\frac 3{1+x}\right)^3$$Cube root both sides and set $x=\sqrt[3]{2}$ gives$$\sqrt[3]{\sqrt[3]2-1}=\frac {3}{1+\sqrt[3]2}=1-\sqrt[3]2+\sqrt[3]4$$Hence$$\sqrt[3]{\sqrt[3]2-1}\color{blue}{=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}}$$A similar technique can be done to show that$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$
