<s>[INST] 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
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$
$$\sqrt[3]{\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$
$$\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}$$
$$\frac{1}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}$$
$${\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}}$$
$$a+b+c=\frac{1}{3}$$
 [/INST] Hint.
$(a,b,c) = \Bigl(\dfrac{1}{9},-\dfrac{2}{9},\dfrac{4}{9}\Bigr)$ $-$  one of rational solutions (ignoring permutations).
So, $a+b+c=\dfrac{1}{3}$. 
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