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?
 A: 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*}
A: 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$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$
A: 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 $c=\frac{1}{9}$
A: 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
$$\sqrt[3]{21\sqrt[3]{6}-17}=\sqrt[3]{18}+\sqrt[3]{4}-\sqrt[3]{3}$$
