Kimchiks926 wrote: Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy equation: $$ f(x^3+y^3) =f(x^3) + 3x^2f(x)f(y) + 3f(x)f(y)^2 + y^6f(y) $$for all reals $x,y$ Putting $x=y=0$ we find that $f(0)=0$. Putting $x=0$ we find that $f(y^3)=y^6f(y)$. Now switching the roles of $x$ and $y$ we find that \[f(x)f(y)(x^2+f(y)-y^2-f(x))=0.\]Clearly $f(x)=0$ is a solution. Otherwise, assume that $f(c) \ne 0$ for some $c$. Then putting $y=c$ we find that for each $x$, either $f(x)=0$ or $f(x)-x^2=f(c)-c^2=:d$ i.e. either $f(x)=0$ or $f(x)=x^2+d$. Suppose that $f(x) \ne 0$. Then $f(x^3) \ne 0$ and hence $x^6+d=x^6(x^2+d)$. This is a contradiction for all but finitely many possible values of $x$. So $f(x)=0$ for all but finitely many values of $x$. But since for $x \ne 0$ we have that $f(x)=0$ is equivalent to $f(x^3)=0$, we can only have $f(x) \ne 0$ for $x \in \{1,-1\}$. Hence $f(x)=0$ for all $x \ne \pm 1$. Putting $y=42$ we find that $f(x^3+74088)=f(x^3)$ for all $x$ and hence $f(x)=0$ for all $x$. Hence the only solution is $\boxed{f(x)=0}$ for all $x$.

Also, there is a similarity with Estonian IMO TST 2020 Problem 3.

Oh, really? I thought it was much more natural to ask the same question with $3x^3f(x)f(y)$ instead of $3x^2f(x)f(y)$...

Tintarn wrote: I thought it was much more natural to ask the same question with $3x^3f(x)f(y)$ instead of $3x^2f(x)f(y)$... Yep, Estonian IMO TST 2020 P3 had $3x^3f(x)f(y)$ instead of $3x^2f(x)f(y)$.

Oh yeah my bad, anyway, both problems are nice.

Let $P(x,y)$ denote the given assertion. We claim that the only solution is the zero function, which clearly works. $P(0,0)$ implies $f(0)=0.$ And $P(0,x)$ implies $f(x^3)=x^6f(x).$ Subtracting $P(x,y)$ from $P(y,x)$ gives $f(x)f(y)(x^2-f(x)-y^2+f(y))=0.$ If $f(x),f(y)\neq 0$ for some $x,y$, then $f(x)=x^2+k.$ So $x^6(x^2+k)=f(x^3)=x^6+k$, which means $x\in \{-1,1\}.$ Therefore, $f(x)=0$ for all $x\notin \{-1,1\}.$ But take $y\neq -1,0,1$ in $P(1,y)$ and $P(-1,y)$ then $f(1)=f(-1)=0$ contradiction.