Assume the common remainder is $k$. This means that $x,y,z$ are the three roots of $t^3-k\equiv 0\pmod{p}$. By Vieta, we get $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)\equiv x+y+z\equiv 0\pmod{p}$. Also we must have $0<x+y+z<3p$ meaning $x+y+z\in\{p,2p\}$ while $x^2+y^2+z^2=pk$ for some positive integer $k$. If $x+y+z=p$ the conclusion is obvious. If $x+y+z=2p$ (note that by the condition we have $p\geq z+1\geq y+2\geq x+3\geq 4>2$) we note by FLT that $x^2+y^2+z^2\equiv x+y+z\pmod{2}$ and so $2p|pk$ and we are done.

$x^3 \equiv y^3 \pmod{p}$ so $y^3-x^3 \equiv 0 \pmod{p}$. By difference of cubes $(y-x)(y^2+xy+x^2) \equiv 0 \pmod{p}$, but since $p>y>x$, $y-x \not\equiv 0 \pmod{p}$ and thus $y^2+xy+x^2 \equiv 0 \pmod{p}$. Similarly $y^2+yz+z^2 \equiv 0 \pmod{p}$ and $z^2+xz+x^2 \equiv 0 \pmod{p}$. Subtracting the first two equations gives $z^2-x^2+yz-xy \equiv 0 \pmod{p}$, or $(z-x)(z+x+y) \equiv \mod{p}$. But similar to before, $z-x \not\equiv 0 \pmod{p}$ and thus $x+y+z \equiv 0 \pmod{p}$. Since $y^2+xy+x^2 \equiv 0 \pmod{p}$, $(x+y)^2 \equiv xy \pmod{p}$. But $(x+y)^2 \equiv (-z)^2 \equiv z^2 \pmod{p}$ so thus $z^2 \equiv xy \pmod{p}$ and thus $x^2+xy+y^2 \equiv x^2 + y^2 + z^2 \equiv 0 \pmod {p}$. The rest of the solution follows according to above.