Baltic Way 2019

Answer: $f(x)=0$ and $f(x)=x^3$.

Substituting $x=y=0$ into the original equation, we obtain $3(f(0){)}^3=0$ which implies $$f(0)=0.(\text{Eq-1})$$ Substituting $x=0$ into the original equation and applying (Eq-1), we get $$f(y^3)=y^{6}f(y).(\text{Eq-2})$$ Replacing the first term of the r.h.s. of the original equation by (Eq-2), we find $$f(x^3+y^3)=x^{6}f(x)+3x^{3}f(x)f(y)+3f(x)(f(y){)}^2+y^{6}f(y).(\text{Eq-3})$$ Swapping the variables $x$ and $y$ in (Eq-3), we find $$3x^{3}f(x)f(y)+3f(x)(f(y){)}^2=3y^{3}f(y)f(x)+3f(y)(f(x){)}^2.(\text{Eq-4})$$ By collecting similar terms in (Eq-4) and factorizing, we get $$f(x)f(y)(f(x)-x^3-f(y)+y^3)=0.(\text{Eq-5})$$ Suppose that $f(x)=f(y)$ for some $x\ne y$. Then $x^3\ne y^3$ as cubing is injective, whence $f(x)-x^3-f(y)+y^3\ne 0$. By (Eq-5), we get $f(x)f(y)=0$, which implies $f(x)=f(y)=0$ by the choice of $x$ and $y$. Consequently, $f$ cannot take non-zero values more than once.

Suppose now that $f(b)=0$ for some $b\ne 0$. Substituting $y=b$ into the original equation, we obtain $f(x^3+b^3)=f(x^3)$ for all $x$. Now if $f(a)\ne 0$ for some $a$ then taking $x=\sqrt[3]{a}$ in the last equality contradicts the previous paragraph since $b^3\ne 0$. Hence $f(x)=0$ for all $x$. This function satisfies the original equation.

It remains to consider the case where $f(x)=0$ implies $x=0$. Substituting $y=1$ into (Eq-5), we obtain $f(x)-x^3-f(1)+1=0$ for all non-zero $x$. In other words, $f(x)=x^3+c$ for all $x\ne 0$ where $c=f(1)-1$. After substituting $y=x$ into (Eq-2), simplifying now gives $c=x^{6}c$ which is possible only if $c=0$. Hence $f(x)=x^3$ for all $x$. This function also satisfies the original equation.