$a=b=c=1$ is counterexample?

DottedCaculator wrote: $a=b=c=1$ is counterexample? oops, forgot to say that $a,b,c$ are different two by two

Cute As Dotted stated above, $a=b=c=1$ is not allowed by the problem statement, so $0 < a + b + c < p$, and in particular $a + b + c$ is nonzero modulo $p$. Thus, $a$, $b$, and $c$ are the roots of some polynomial $x^3 - mx + n$ in $\mathbb F_p[x]$, where $m$ is nonzero. But the function $x\mapsto mx - n$ is bijective modulo $p$, so $a^3$, $b^3$, and $c^3$ are distinct mod $p$. $\square$ Remark. I was originally going to type something much more complicated for the argument in the first sentence, but I realized I could make it much simpler while typing!

Let $a>b>c$ without loss; assume $a^3\equiv b^3\pmod{p}$. Then $p\mid (a-b)(a^2+ab+b^2)$ implies, together with $0<a-b<p$, that $p\mid a^2+ab+b^2$. Adding $p=ab+bc+ca$ to this, we get $p\mid (a+b)^2 + c(a+b)=(a+b)(a+b+c)$. But since $0<a+b<a+b+c<p$, this is a contradiction. The case $a^3\equiv c^3\pmod{p}$ is handled analogously and $b^3\equiv c^3\pmod{p}$ is impossible as $a>b>c$.

djmathman wrote: Thus, $a$, $b$, and $c$ are the roots of some polynomial $x^3 - mx + n$ in $\mathbb F_p[x]$, where $m$ is nonzero. I think it should be $x^3-mx^2+n$ which makes the rest of the argument slightly more complicated, but it still works of course.