, 2024

Answer: $f(x) = 0$ and $f(x) = x$.

$f(x + y)^3 = (x + 2y)f(x^2) + f(f(y))(x^2 + 3xy + y^2)$.

By putting $(x, y) = (0, 0)$ to the original equation we get $f(0) = 0$.

By putting $(x, y) = (x, 0)$ to the original equation we get $f(x)^3 = x f(x^2)$.

By putting $(x, y) = (0, x)$ to the original equation we get $f(x)^3 = f(f(x)) x^2$.

Replace this in the original equation to get
$$f(x + y)^3 = (x^2 + 2xy)f(f(x)) + (x^2 + 3xy + y^2)f(f(y)).$$
Reversing the $x, y$ order, we get
$$f(x + y)^3 = (x^2 + 3xy + y^2)f(f(x)) + (2xy + y^2)f(f(y)).$$
Since the RHS in both expressions are equal we get
$$f(f(x))(x + y)y = f(f(y))(x + y)x$$
for all real numbers $x, y$.

Setting $y = 1$ and letting $f(f(1)) = k$, we have $f(f(x)) = kx$ for every $x \in \mathbb{R} \setminus \{-1\}$, setting $y = 2$ and $x = -1$ gives that this is true for all real numbers.

Putting this into $f(x)^3 = f(f(x))x^2$, we get $f(x) = cx$ for a constant $c = \sqrt[3]{k}$, and putting this into $x f(x^2) = x^2 f(f(x))$ we get $c = 0$ or $c = 1$.

Therefore, the solutions are $f(x) = 0$ and $f(x) = x$ and they satisfy the original equation.