TITLE: Fermat's Last Theorem (Case n = 3) Question
QUESTION [25 upvotes]: A very simple question.
We all know that there are no solutions to $x^3 + y^3 = z^3$ for integer $x$, $y$ and $z$, $xyz\neq 0$, but are rational $x$, $y$ and $z$ possible?  Thanks.

REPLY [39 votes]: No, because if $x=\frac{a}{b}$, $y=\frac{c}{d}$, and $z=\frac{e}{f}$ were rational solutions, so that
$$\left(\frac{a}{b}\right)^3+\left(\frac{c}{d}\right)^3=\left(\frac{e}{f}\right)^3,$$
then there would be an integer solution
$$(adf)^3+(bcf)^3=(bde)^3$$
(edit: this is just writing out the details of Did's answer)