$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$ has infinitely many non-trivial solutions Trying to solve this I find out the following problem in which it is not necessary the condition $x^3=y^3=z^3$ in some $\mathbb F_p$:
Prove there are infinitely many pairwise coprime triples of distinct natural numbers, $(x,y,z)$, such that:
$$\frac{x^3+y^3+z^3}{x+y+z}\in \mathbb N$$

Yes, there are infinitely many pairwise coprime solutions $x,y,z$.
$(x,y,z)=(m,n,m+n)$ with $\gcd(m,n)=1$ and $m$ even.
Motivation: $$x^3+y^3+z^3=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz$$
Therefore:
$$x+y+z\mid x^3+y^3+z^3\iff x+y+z\mid 3xyz$$
If $(x,y,z)=(m,n,m+n)$, then $$\iff 2(m+n)\mid 3mn(m+n)\iff 2\mid mn$$
Since you want pairwise coprime solutions, we can let $\gcd(m,n)=1$.
