$p|x^3-y^3$ and $0<x<y<p$,so $p|x^2+xy+y^2$,also $p|x^2+xz+z^2$,then $p|(y-z)(x+y+z)$,we get $p|x+y+z$.With $0<x+y+z<3p$,$x+y+z=p$ or $2p$. We only need to prove $p|xy+yz+zx$,it's true because $p|\sum_{cyc}(x^2+xy+y^2)-2(x+y+z)^2$.

I'm happy to see a person with a username similar to mine. The complex number analogue is obvious. Suppose $x, y, z$ are distinct complex numbers. If $x^3, y^3, z^3$ are equal, prove that if $x + y + z=0$ then $x ^ 2 + y ^ 2 + z ^ 2=0$. Just do the analogue in the finite field. nguyenalex wrote: Suppose $p$ is a prime number and $x, y, z$ are integers satisfying $0 < x < y < z <p$. If $x^3, y^3, z^3$ have equal remainders when divided by $p$, prove that $x ^ 2 + y ^ 2 + z ^ 2$ is divisible by $x + y + z$.