Find all functions $f : \mathbb R \to \mathbb R$ such that: $f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)$

Question

Here is the question:

Find all functions $f : \mathbb R \to \mathbb R$ such that: $$f\left(x^3\right)+f\left(y^3\right) = (x + y)f\left(x^2\right)+f\left(y^2\right)-f(xy)$$

What I tried:

Note that the function $f\left(x^3\right)+f\left(y^3\right)$ is symmetric. From here we get $$(x + y)f\left(x^2\right) + f\left(y^2\right) - f(xy) = f(x + y)f\left(y^2\right) + f\left(x^2\right) - f(xy)$$ which gives $(x + y - 1)f\left(y^2\right) = (x + y - 1)f\left(x^2\right)$. From here I cannot proceed further. I could have cancelled $(x + y - 1)$ but I haven't proved that $(x + y - 1) \neq 0$ and neither I don't know how to proceed with $\left(y^2\right) = f\left(x^2\right)$ even if we can cancel $x + y - 1$. Any hints or suggestions will be greatly appreciated.

Answer 1

Here's how to get $f(x^2) = f(y^2)$.

If $x+y-1 \neq 0$, we must have $f(x^2) = f(y^2)$. F any $x^2$ and $y^2$, pick a $z$ such that $x+z-1$ and $y+z-1$ are nonzero, which is always possible by taking $z \neq 1-x, 1-y$. Then $f(x^2) = f(z^2) = f(y^2)$. So $f(x^2) = f(y^2)$ for all reals $x, y$, and hence $f$ is constant on nonnegative reals.