WLOG, we can just assume that $x, y, z$ are distinct positive integers less than $p$. $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\} $ means the following $x^3\equiv y^3\equiv z^3 \pmod{p}$. Thus, $(x^{-1}y)^3\equiv (x^{-1}z)^3\equiv 1 \pmod{p}$. Thus, for the primitive root $g$, we can say the following without loss of generality. $x^{-1}y \equiv g^{\frac{p-1}{3}}$, $x^{-1}z \equiv g^{\frac{2(p-1)}{3}} \pmod{p}$. Thus, $y\equiv xg^{\frac{p-1}{3}}, z\equiv g^{\frac{2(p-1)}{3}} \pmod{p}$. Meanwhile, considering $(g^{\frac{p-1}{3}}-1)(g^{\frac{2(p-1)}{3}}+g^{\frac{p-1}{3}}+1)\equiv g^(p-1)-1\equiv 0 \pmod{p}$ and $g^{\frac{p-1}{3}}\not\equiv 1 \pmod{p}$, we can make the conclusion like this. $x+y+z\equiv 0 \pmod {p}$ Thus, $x+y+z=p _{or} 2p$. Therefore, everything we have to do is to prove the following $x^5+y^5+z^5\equiv 0 \pmod{p}$. However, considering $k^2+k+1 \mid k^{10}+k^5+1$ it is easy to prove. Done

It's exactly Poland 2003: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2070729&sid=64df69e2c903958785a9c858b780a3de#p2070729 just in the end you use that $x^5+y^5+z^5\equiv x^3(x^2+y^2+z^2)$.

We have $x^3\equiv y^3\equiv z^3\pmod p$. Note that $p|y^3-x^3$, and $0<y-x<p$, so $p|x^2+xy+y^2$. Similarly, $p|y^2+yz+z^2$ and $p|z^2+zx+x^2$. We see $p|x^2-y^2+xz-yz$, so $p|(x-y)(x+y+z)$. Hence, $p|x+y+z$. Therefore, $x+y+z$ is either $p$ or $2p$. Note that $p|x^2+y^2+z^2+2(xy+yz+zx)$ and $p|2(x^2+y^2+z^2)+xy+yz+zx$. Since $p>3$, we must have $p|x^2+y^2+z^2$. Hence, $x^5+y^5+z^5\equiv x^3(x^2+y^2+z^2) \equiv 0 \pmod p$. Also, $x^5+y^5+z^5\equiv x+y+z\pmod 2$. Therefore, $x+y+z|x^5+y^5+z^5$.