9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
8
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$

$\endgroup$
3
$\begingroup$

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 $y^3+1=3=(y+1)(y^2-y+1)$

Hence

$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}=\frac{1}{9}(y^2-y+1)$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.