Legendre conjectured that for $p, q, r, s \in \mathbb{N}$ it is impossible that $$\biggl(\frac{p}{q}\biggl)^3+\biggl(\frac{r}{s}\biggl)^3=6$$ for any $p, q, r, s$.
Eventually Henry Ernest Dudeny found a counterexample, where $p=17, q=21, r=37, s=21$: $$\biggl(\frac{17}{21}\biggl)^3+\biggl(\frac{37}{21}\biggl)^3=\frac{4913}{9261}+\frac{50653}{9261}=6.$$
Does the original conjecture hold up if we assume that none of $p, q, r, s$ repeat? That is, none of the four numbers can equal each other?
The given equation is equivalent to $$p^3s^3+q^3r^3=6q^3s^3$$ If we assume that $p$ and $q$ are coprime as well as $r$ and $s$ (otherwise the counterexample can easily be modified such that all numbers are distinct) , we get $q^3|s^3$ because of $q^3|p^3s^3$ and $s^3|q^3$ because of $s^3|q^3r^3$. This implies $s^3=q^3$ and therefore $s=q$