Note: $f^3(x)=f(x)^3$.
Question:
Find all the functions $f:\mathbb R\to\mathbb R$ satisfy $\forall x,y,z\in\mathbb R, x+y+z=0$, we have $f(x^3)+f^3(y)+f^3(z)=3xyz$.
My ideas:
It is obvious that $f(x)=x$ is a solution. It is likely that $f(x)=x$ is the only solution.
Let $x=0$, then $y=0$, we can find that $f(x^3)=f^3(x)$ for all $x\in\mathbb R$.
So $f(x^3)+f(y^3)+f(z^3)=3xyz=x^3+y^3+z^3$ for all $x,y,z\in\mathbb R, x+y+z=0$.I think it is related to Cauchy's Functional Equation, which needs $f(x+y)=f(x)+f(y)\ \forall x,y\in\mathbb R$. However, I can't prove it.
Maybe it is useful to prove $f$ is injective, but I have no idea.
I will show how we transform this into a Cauchy's functional equation (since this is what the poster attempts to do).
Take $x=y=z=0$, then $f(0)+2f(0)^3=0$. The only real solution is $f(0)=0$.
Take $x=0,z=-y$, then $f(y)^3+f(-y)^3=0$ for any $y\in\mathbb{R}$. Thus $(-f(y))^3=f(-y)^3$ and hence $-f(y)=f(-y)$. This shows that $f$ is an odd function.
Take $z=0,y=-x$, then $f(x^3)+f(-x)^3=0$ for any $x\in\mathbb{R}$. Thus $f(x^3)=f(x)^3$ since $f$ is an odd function.
To sum up we have $f(x^3)+f(y^3)+f(z^3)=3xyz$ for $x+y+z=0$. In the following we need the identity $x^3+y^3+z^3=3xyz$ for $x+y+z=0$, which is simply because $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Now define $g(t)=f(t^3)-t^3$. Then $g$ is also an odd function and $g(0)=0$. We have $$g(x)+g(y)+g(z)=f(x^3)+f(y^3)+f(z^3)-(x^3+y^3+z^3)=0$$ if $x+y+z=0$. Therefore for $x,y\in\mathbb{R}$, we have $g(x)+g(y)=-g(-x-y)=g(x+y)$, a Cauchy's functional equation.