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$\textbf{Problem} $ Find rational numbers $a,b,c$ satisfying \begin{align*} (2^\frac13-1)^\frac13 = a^\frac13+b^\frac13+c^\frac13 \end{align*}

My Attempt: I try to $2^\frac13-1 = (a^\frac13+b^\frac13+c^\frac13)^3$ and compare with rational numbers and irrational numbers in LHS and RHS.

Any help is appreciated...

Thank you!

Update: I found the answer. I want to find rational numbers $a,b,c$ without assumption $a=1/9,b=-2/9,c=4/9$. Thus, I found the identity: $$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$

1) How to get the identity?

2) I want to know about uniqueness $(a,b,c)$ satisfying $(2^\frac13-1)^\frac13=a^\frac13+b^\frac13+c^\frac13$

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    Do you have reason to think that there are such rational numbers $a,b,c$ ? – Lubin Jul 11 '18 at 13:19
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    Possible duplicate of Ramanujan's Surd : Proof required. – Ng Chung Tak Jul 11 '18 at 13:52
  • How to get the identity : $$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$??? –  Jul 11 '18 at 14:26
  • I want to find a,b,c without assumption $a=1/9,b=-2/9,c=4/9$... Thus, I found the identity. Do you know how to get that? –  Jul 11 '18 at 14:36
  • @w.sdka, I think you should stress your question 2), asking if $(a,b,c)=(-2/9,1/9,4/9)$ is the only solution with $a\le b\le c$, since that does not seem to be addressed elsewhere. Also, what does "How to get the identity?" mean? It looks to me like you can verify it by simply cubing both sides and sorting stuff out. – Barry Cipra Jul 11 '18 at 15:02

1 Answers1

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Set $t:=\sqrt[3]{2}$. Then, $t^3=2$ and $t^3-1=1$, whence $$t-1=\frac{t^3-1}{t^2+t+1}=\frac{1}{t^2+t+1}=\frac{3}{3t^2+3t+3}=\frac{3}{t^3+3t^2+3t+1}=\frac{3}{(t+1)^3}\,.$$ Therefore, $$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}}{t+1}\,.$$ Now, $t^3+1=3$, so $$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{t^3+1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{3}=\frac{t^2-t+1}{\sqrt[3]{9}}\,.$$ Plugging in $t=\sqrt[3]{2}$, we obtain $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{4}{9}}+\sqrt[3]{-\frac{2}{9}}+\sqrt[3]{\frac19}\,.$$

In fact, this is the only possible way to write $\sqrt[3]{\sqrt[3]{2}-1}=\frac{1}{3}\sqrt[3]{12}-\frac{1}{3}\sqrt[3]{6}+\frac{1}{3}\sqrt[3]{3}$ as a sum $$\sum_{j=1}^n\,x_j\,\sqrt[r_j]{d_j}\,,$$ where $x_1,x_2,\ldots,x_n\in\mathbb{Q}\setminus\{0\}$, $r_1,r_2,\ldots,r_n\in\mathbb{Z}_{>0}$, and for $j=1,2,\ldots,n$, each $d_j\neq0$ is an $r_j$-power-free integer (with $d_j=1$ iff $r_j=1$, and with $d_j>1$ if $r_j>1$ is odd). This is because radicals are linearly independent over $\mathbb{Q}$. See a proof here. (As a consequence, $a$, $b$, and $c$ are unique, up to permutation.)

Batominovski
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