I was trying to solve this functional equation that I found in some papers. $$f:\mathbb{R} \to \mathbb{R}$$ $$2f(x+y)+6y^3=f(x+2y)+x^3$$ for every $x,y \in \mathbb{R}$
First, I made $x=y=0$. Then: $$2f(0)+0=f(0)+0$$ $$f(0)=0$$
I then made $x=-y$. I get: $$2f(-y+y)+6y^3=f(-y+2y)+(-y)^3$$ $$2f(0)+7y^3=f(y)$$ Because we know that $f(0)=0$, we can say that: $$f(x)=7x^3$$
But when I try to use $f(x)=7x^3$ in the original functional equation I don't get something that is equal for every $x,y$. Does that mean there are no solutions to this functional equation or did I do something wrong?
Can we make this a sensible problem? Maybe a functional equation not over the reals, but over some other field? We already have $7x^3 = x^3$, so $6=0$, so the characteristic is either $2$ or $3$.
Yes. In fact, $f(x) = x^3$ does satisfy this functional equation over any field of characteristic $2$ or $3$.