Rationality of an expression

[jnml]

Given $$x \in \mathbb{Q}, x \neq 0,$$ can or cannot $$\sqrt[3]{x^3+1} \in \mathbb{Q}$$ be true?

[paulunderwood]

Let x=1, then 2^{1/3} is irrational.

FLT might have something to do with the stated problem.

[axn]

fermat's last theorem

EDIT:- Loophole abuse - x=-1

[LaurV]

Say \(x\in\mathbb{Q}\) not 0, then \(x\) can be written \(\frac ab\), so you have \(\ ^3\sqrt{(\frac ab)^3+1}=\ ^3\sqrt{\frac{a^3}{b^3}+1}=\ ^3\sqrt{\frac{a^3+b^3}{b^3}} = \frac{\ ^3\sqrt{a^3+b^3}}{b}\) which is rational, so it is \(=\frac mn\), therefore \(\ ^3\sqrt{a^3+b^3}=\frac{bm}{n}\), and renaming the last with \(c\), we have \(\ ^3\sqrt{a^3+b^3}=c\) or \(a^3+b^3=c^3\) and as Paul said, we just found a counterexample for FLT.



Therefore, that can never be true.

[R. Gerbicz]

Originally Posted by axn

EDIT:- Loophole abuse - x=-1

and there is no more solution.

[jnml]

Originally Posted by LaurV

... which is rational ...

Did you mean irrational?

[paulunderwood]

Originally Posted by jnml

Did you mean irrational?

It is presumed rational until the contradiction by FLT is established.

[jnml]

Originally Posted by paulunderwood

It is presumed rational until the contradiction by FLT is established.

I see, thanks. I did not assume that particular implicit assumption ;-)