20180906, 13:02  #1 
Feb 2012
Prague, Czech Republ
11001001_{2} Posts 
Rationality of an expression
Given $$x \in \mathbb{Q}, x \neq 0,$$ can or cannot $$\sqrt[3]{x^3+1} \in \mathbb{Q}$$ be true?

20180906, 13:12  #2 
Sep 2002
Database er0rr
67^{2} Posts 
Let x=1, then 2^{1/3} is irrational.
FLT might have something to do with the stated problem. Last fiddled with by paulunderwood on 20180906 at 13:18 
20180906, 13:48  #3 
Jun 2003
2^{2}·3^{2}·151 Posts 
fermat's last theorem
EDIT: Loophole abuse  x=1 Last fiddled with by axn on 20180906 at 14:00 
20180906, 14:51  #4 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10100000101001_{2} Posts 
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. Last fiddled with by LaurV on 20180906 at 14:53 
20180906, 15:13  #5 
"Robert Gerbicz"
Oct 2005
Hungary
1611_{10} Posts 

20180906, 15:27  #6 
Feb 2012
Prague, Czech Republ
3·67 Posts 

20180906, 15:47  #7 
Sep 2002
Database er0rr
10611_{8} Posts 

20180906, 15:57  #8 
Feb 2012
Prague, Czech Republ
3×67 Posts 

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