Rationalizing expressions

In my precalc book, I have the following problem:

Calculate $a+b+c$ if $a,b,c\in\mathbb{Q}$ and $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$$

I think that the RHS can stay untouched, while operating the LHS, but I can't find a way to factor $\sqrt[3]{2}-1$ as the third power of something. Any help is greatly appreciated.

With the help of Olegg, i got the solution $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$ $$\sqrt[3]{\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$ $$\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}$$ $$\frac{1}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}$$ $${\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}}$$ $$a+b+c=\frac{1}{3}$$

-
do you mean "calculate $a + b + c$ if..."? –  James Jun 15 '13 at 15:24
Edited: Rephrased the question for better understanding –  chubakueno Jun 15 '13 at 16:43

$(a,b,c) = \Bigl(\dfrac{1}{9},-\dfrac{2}{9},\dfrac{4}{9}\Bigr)$ $-$ one of rational solutions (ignoring permutations).
So, $a+b+c=\dfrac{1}{3}$.
One of the things I can notice in your solution would be that it follows the form $x^2-xy+y^2$, I think i will try to backtrack from there. –  chubakueno Jun 15 '13 at 17:48
Yes, @chubakueno, it was simple computer search. But I have small doubts: is it unique triple? For example, $\sqrt[3]{1}-\sqrt[3]{2}+\sqrt[3]{8} = \sqrt[3]{27}+\sqrt[3]{16}-\sqrt[3]{54}$. How to show uniqueness? –  Oleg567 Jun 15 '13 at 22:20
I think that a proof would start showing that a radical has a unique representation as an irreducible radical and showing that if two irreducible radicals dont have the same base, they cannot be expressed in only one, then using that property as a tool for demonstranting that if $x\sqrt[n]{a}$, and $y\sqrt[n]{b}$ are irreducible radicals, then is necessary that if$a \neq b$ and $x\sqrt[n]{a}+y\sqrt[n]{b}=z\sqrt[n]{c}+w\sqrt[n]{d}$, then {$z\sqrt[n]{c}, w\sqrt[n]{d}$} must be a permutation of {$x\sqrt[n]{a}$,$y\sqrt[n]{b}$} Then using this in a general case for any number of cases an $n$. –  chubakueno Jun 17 '13 at 6:24