I try as following let \begin{eqnarray} x= \sqrt[3]{a} \\ y= \sqrt[3]{b} \\ z= \sqrt[3]{c} \\ x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\ \end{eqnarray} We know that \begin{equation} x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z) \end{equation} that turns out to be \begin{equation} (x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z) \end{equation} plug in $x+y+z$, we will get \begin{equation} \sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z) \end{equation} From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?

  • $\begingroup$ MathJax works in the title, don't you know? $\endgroup$ – Shaun Mar 25 '18 at 13:50
  • $\begingroup$ Thanks for the correction of my title of my question. $\endgroup$ – Wuttipong Kumwilaisak Mar 25 '18 at 14:24
  • $\begingroup$ I don't see how to get $3(x+y)(x+z)(y+z)$, did you put additional constraints on the $x,y,z$? $\endgroup$ – punctured dusk Mar 25 '18 at 14:34
  • $\begingroup$ $a$,$b$, and $c$ are just rational number. $\endgroup$ – Wuttipong Kumwilaisak Mar 25 '18 at 14:41
  • $\begingroup$ $(2^{1/3}-1)^{1/3}=\frac1{a^{1/3}}(1-2^{1/3}+4^{1/3})$ is pre-test of Japanese math olympiad in 1995. $\endgroup$ – Takahiro Waki Mar 25 '18 at 17:25

An observation:

\begin{align*} (\sqrt[3]{4}-\sqrt[3]{2}+1)^3&=\frac{(\sqrt[3]{8}+1)^3}{(\sqrt[3]{2}+1)^3}\\ &=\frac{27}{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}\\ &=\frac{9}{\sqrt[3]{4}+\sqrt[3]{2}+1}\\ &=\frac{9(\sqrt[3]{2}-1)}{\sqrt[3]{8}-1}\\ &=9(\sqrt[3]{2}-1)\\ \left(\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{-2}{9}}+\sqrt[3]{\frac{1}{9}}\right)^3&=\sqrt[3]{2}-1 \end{align*}

  • $\begingroup$ Thank you so much but how you can get this observation? $\endgroup$ – Wuttipong Kumwilaisak Mar 26 '18 at 1:12
  • $\begingroup$ I just know that this problem partly came from Ramanujan's identity. How we systematically solve this kind of problem? $\endgroup$ – Wuttipong Kumwilaisak Mar 26 '18 at 1:14
  • $\begingroup$ I don’t have a systematic approach yet. I just think that the cube of the RHS is complicated and with many different cube roots. Maybe $a,b,c$ are related. $p\sqrt[3]{4}+q\sqrt[3]{2}+r$ is a natural choice. $\endgroup$ – CY Aries Mar 26 '18 at 1:22

Here is an alternative to denesting $(2^{1/3}-1)^{1/3}$. First, set $x^3=2$ so that$$x^3-1=1$$Factoring the left-hand side and isolating $x-1$ gives$$x-1=\frac 1{1+x+x^2}=\frac 3{3+x+3x^2}=\frac 3{(1+x)^3}$$Multiply both sides by $9$ to complete the cube$$9(x-1)=\left(\frac 3{1+x}\right)^3$$Cube root both sides and set $x=\sqrt[3]{2}$ gives$$\sqrt[3]{\sqrt[3]2-1}=\frac {3}{1+\sqrt[3]2}=1-\sqrt[3]2+\sqrt[3]4$$Hence$$\sqrt[3]{\sqrt[3]2-1}\color{blue}{=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}}$$A similar technique can be done to show that$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$


I try this: $y=\sqrt[3]{2}$,$y^3=2$, $x= \sqrt[3]{\sqrt[3]{2}-1}$ Therefore, $x^3=y-1$ $y^3-1=(y-1)(y^2+y+1)=1$ Because $y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$ and $y^3+1=3=(y+1)(y^2-y+1)$



Therefore, based on the definition of $y$, we can obtain $a=\frac{4}{9}$,$b=-\frac{2}{9}$, and $c=\frac{1}{9}$


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