Finding necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$ My question is that:

Find the necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$.

I know that answer is $k=-3$, as I know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
But I can't prove that $k=-3$ is also a necessary condition. A mathematical proof is needed.
Thanks.

The condition necessary and sufficient for a polynomial $f(x)$ to be divisible by $(x-a)$ is that $f(a)=0$.
Let, $$f(x)=x^3+y^3+z^3+kxyz$$
For this polynomial to be divisible by $x+y+z$, it is necessary and sufficient that $f(-y-z)=0$.
However,
$$f(-y-z)=-(y+z)^3-kyz(y+z)+y^3+z^3=-(k+3)yz(y+z)=0$$
Simplifying it gives us, $k=-3$.
Thus for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$, it is necessary and sufficient that $k=-3$.
