Strategies to find the set of functions $f:\mathbb R\to\mathbb R$ satisfing the functional equation: $f(x^3)+f(y^3)=(x+y)(f(x^2)+f(y^2)-f(xy))$

Question

My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation: $$f\big(x^3\big)+f\big(y^3\big)=(x+y)\Big(f\big(x^2\big)+f\big(y^2\big)-f(xy)\Big):\forall x, y\in\mathbb R$$

I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis.

Thanks in advance.

Answer 1

Here is a typical start. Put $x=y=0$. The right side is $0$, so $f(0)=0$.

Now set $y=0$, and let $x$ roam freely. Since $f(0)=0$, we get $f(x^3)=xf(x^2)$.

Set $y=-x$. We get $f(x^3)+f(-x^3)=0$. Since everything is a cube, we have $f(-u)=-f(u)$ for all $u$.

Now explore $x=y$. Can we learn anything from setting $x$ and/or $y$ qual to $1$?

A little playing has gotten us a lot of information, enough that we should be able to complete things.

Answer 2

If it is of any help the only continuous solutions are of the form $f(x)=xf(1)$. Indeed we can use the property $f(x^3)=xf(x^2)$ to show $$f(x)=x^\frac{1}{3}f(x^\frac{2}{3})=x^\frac{1}{3}x^\frac{2}{9}f(x^\frac{4}{9})=\cdots=x^{\frac13\sum_i(\frac23)^i}f(1)=xf(1).$$ In fact we only need continuity at $1$.