Functional equation $f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$ Find all of the functions defined on the set of integers and receiving the integers value, satisfying the condition:
$$f(a+b)^3-f(a)^3-f(b)^3=3f(a)f(b)f(a+b)$$
for each pair of integers $(a,b)$.
 A: Not cleaned. I just wrote it as I thought about it.
$A^3+B^3+C^3-3ABC=(A+B+C)(A+wB+w^2C)(A+w^2B+wC)$. So in our case
$$\left(f(a+b)-f(a)-f(b)\right)\left(f(a+b)-wf(a)-w^2f(b)\right)\left(f(a+b)-w^2f(a)-wf(b)\right)=0.$$
If for some $a,b$ we have $f(a+b)-wf(a)-w^2f(b)=0$ then $f(a+b)-wf(a)-w^2f(b)$ is divisible by $w^2+w+1$. Using long division we get $f(a+b)-wf(a)-w^2f(b)=(w^2+w+1)(-f(b))+(f(b)-f(a))w+f(a+b)-f(b)$. From where $f(b)=f(a)$ and $f(a+b)=-f(b)$.
So we are getting that, for each $a,b$, either:
$$f(a+b)=f(a)+f(b)$$
or $$f(a)=f(b)=-f(a+b).$$
We clearly get that $f(0)=0$.
If for some $x$, $f(x)=f(-x)$ then we get $f(x)=f(-x)=0$, otherwise we get $f(x)=-f(-x)$. If for some $x$, $f(x)=0$, then $f(nx)=0$.
If $f(1)=0$, then $f(a+1)=f(a)$. So, $f$ is constant equal to zero.
If $f(1)=r\neq0$ then $$f(a+1)=f(a)+r$$. Therefore $f(x)=rx$.
A: An easier answer.
Using $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ and putting $x=f(a+b),y=-f(a),z=-f(b)$ we get that $x^3+y^3+z^3-3xyz=0$ so we have that either $x+y+z=0$ or $x^2+y^2+z^2-xy-yz-zx=0$. If $x+y+z=0$ then we have that $f(a+b)=f(a)+f(b)$ which gives us $f(a)=af(1)$. If the second one is true we get from it that $x=y=z$, which means that $-f(a+b)=f(a)=f(b)$. From which we can get that $f(x)=0$ fot evey real number x. So those are the two solutions.
