Solve $z^3=(sr)^3$ where $r,s,z$ are integers? Let $x,y,z$ be 3 non-zero integers defined as followed: 
$$(x+y)(x^2-xy+y^2)=z^3$$
Let assume that $(x+y)$ and $(x^2-xy+y^2)$ are coprime
and set $x+y=r^3$ and $x^2-xy+y^3=s^3$
Can one write that $z=rs$ where $r,s$ are 2 integers?
I am not seeing why not but I want to be sure.

Yes.
If $z^3 = r^3s^3$ we can take cube roots of both sides to get $z = rs$. It's  valid to do this because $n$ is the only solution to $({n^3})^{1/3}$ for real $n$. If we go to complex numbers we have to be more careful because cube roots have 3 solutions.
