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I came across a problem on the web, it is as follows:

Question:If $ x $, $ y $ and $ z $ are rational numbers such that $ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then find $ x,y,z $.

From Wikipedia|Nested Radicals|Some identities by Ramanujan, it is clear that the values of $x,y,\text{and} \ z$ is $\sqrt[3]{\frac{1}{9}}$,$-\sqrt[3]{\frac{2}{9}}$ and $\sqrt[3]{\frac{4}{9}}$ respectively.

However, even after going through the references from the wikipedia and from the web [2], [3], [4]; I still fail at understanding it.

Can someone please show me the proof of this result in an intuitive way?

My approach in attacking this problem is as follows,

Let, $y = \sqrt[3]{\sqrt[3]{2}-1}$

Cubing both sides,

$$ y^{3} = \sqrt[3]{2}-1 \\ \implies y^{3}+1=\sqrt[3]{2} \\ \implies (y^{3}+1)^{3}= 2 $$

Now, using the identity : $(a+b)^3=a^3 + 3a^{2}b + 3ab^{2} + b^{3}$

$$ y^{9} + 3y^{6}+3y^{3}+1 = 2 \\ \implies y^{9} + 3y^{6}+3y^{3} - 1 = 0 $$

Now, I get the cubic equation using the substitution, $y^{3} = m$

$$ m^{3} + 3m^{2}+3m - 1 = 0 $$

Seems like I get lost in my own web.

Can someone help me out?

Pragyaditya Das
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  • Also see https://math.stackexchange.com/questions/871639/denesting-radicals-like-sqrt3-sqrt32-1 – Paramanand Singh Apr 19 '17 at 07:12
  • To prove you can put $x=\sqrt[3]{2}$ and consider $y=(1-x+x^{2})/\sqrt[3]{9}$ and show directly that $y^3=x-1$. – Paramanand Singh Apr 19 '17 at 07:14
  • I guess, I wasn't clear enough. What I want is not a possible trick to the values of $x$, $y$ and $z$. I want a full and detailed mathematical proof of the value obtained by Ramanujan. Also, I don't understand why you consider $y = \frac{1-x+x^2}{\sqrt[3]{9}}$? – Pragyaditya Das Apr 19 '17 at 08:21

1 Answers1

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You want a proof of the identity $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$ which can be written as $$\sqrt[3]{x-1}=\frac{1-x+x^{2}}{\sqrt[3]{9}}$$ where $x=\sqrt[3]{2}$. Thus we let the RHS be denoted by $y$ and show that $y^{3}=x-1$. Since $x, y$ are real this proves the desired identity.

Now we can see that $$y^{3}=\frac{(1-x+x^{2})^{3}}{9}=\frac{(x^{3}+1)^{3}}{9(x+1)^{3}}=\frac{3}{x^{3}+3x^{2}+3x+1}=\frac{1}{1+x+x^{2}}=\frac{x-1}{x^{3}-1}=x-1$$ In the above algebraic manipulation $x^{3}$ has been replaced with $2$.

Ramanujan was expert in algebraic manipulation of numbers as well as symbolic expression so much so as to make MAPLE and MACSYMA almost unnecessary and he obtained most of such denesting of radicals as special cases of general algebraic identities. It was not exactly like solving equations to find value of $x, y,z$ as you ask in the question.

Paramanand Singh
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    Thanks @JyrkiLahtonen. I am a big fan of Ramanujan (you can see some of his work on my blog https://paramanands.blogspot.in/p/archives.html). His mathematical adventures were much more significant compared to other Indian mathematicians. Unfortunately other Indian mathematicians from past are more well known in our country. So I try to use every opportunity to highlight work of Ramanujan. – Paramanand Singh Apr 19 '17 at 09:39
  • Thank you for the great explanation. And it is great to know a man like Ramanujan. :) Blessed. – Pragyaditya Das Apr 20 '17 at 05:58