Rationalizing expressions

In my precalc book, I have the following problem:

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

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

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

• 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, $\sqrt{1}-\sqrt{2}+\sqrt{8} = \sqrt{27}+\sqrt{16}-\sqrt{54}$. 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