# 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.

• We had a jam session with the $k=-3$ case a few years ago. Fun reading IMHO. Some of those answers can be made to work here. Not calling this a dupe. I just feel that these two threads can be read together. Dec 30, 2016 at 11:56
• Since $x^3+y^3+z^3-3xyz$ is divisible by $x+y+z$, $x^3+y^3+z^3+kxyz$ is divisible by $x+y+z$ if and only if their difference $(k+3)xyz$ is divisible by $x+y+z$, hence, either $k+3=0$ or $xyz$ is divisible by $x+y+z$. But $xyz$ is not divisible by $x+y+z$ hence one is left with the condition that $k+3=0$, qed.
– Did
Dec 30, 2016 at 11:58

A low key answer is: We have $x^3+y^3+z^3+kxyz = (3+k)xyz + (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Thus this means $(3+k)xyz$ must be divisible to $x+y+z$ for all $x,y,z$ integers. Put $x = y = 1$ we have $(3+k)z$ is divisible to $z+2$, thus you can write $(3+k)z = n(z+2)$, for all $z$ integer. Thus $n = \dfrac{(k+3)z}{z+2} = k+3 - \dfrac{2k+6}{z+2} \in \mathbb{Z}$ for all $z$ integers. Thus $2k+6 = 0 \implies k = -3$ as claimed.

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 $k=-3$.

• To be divisible by $x+y=z$??? Please check again and edit.
– user371838
Dec 30, 2016 at 11:55