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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 $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$ $$\sqrt[3]{\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}}$$ $$\sqrt[3]{\frac{1}{(\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}})^3}}$$ $$\frac{1}{\sqrt[3]{\frac{1}{3}}+\sqrt[3]{\frac{2}{3}}}$$ $${\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}}$$ $$a+b+c=\frac{1}{3}$$

chubakueno
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1 Answers1

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Hint.

$(a,b,c) = \Bigl(\dfrac{1}{9},-\dfrac{2}{9},\dfrac{4}{9}\Bigr)$ $-$ one of rational solutions (ignoring permutations).

So, $a+b+c=\dfrac{1}{3}$.

Oleg567
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  • But how did you came up with this values? I mean, a computer search with denominators and numerators from -20 to 1 and 1 to 20 could give me this values, but I would like to justify it. Anyways, thanks in advance. – chubakueno Jun 15 '13 at 17:40
  • 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
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    Coming up with such solutions might be hard, since this identity is commonly attributed to Ramanujan; see this paper – Ian Mateus Jun 15 '13 at 19:10
  • 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 06:24
  • Thank you so much, @chubakueno. – Oleg567 Jun 17 '13 at 20:20