Problem 3. Let $\mathbb{R}$ denote the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f\left(x^{3}\right)+f(y)^{3}+f(z)^{3}=3 x y z
$$

for all $x, y, z \in \mathbb{R}$ such that $x+y+z=0$.

(Kyprianos-Iason Prodromidis)
First Solution. First of all, for $x=y=z=0$, we get that

$$
f(0)+2 f(0)^{3}=0 \Leftrightarrow f(0)\left(2 f(0)^{2}+1\right)=0 \Leftrightarrow f(0)=0
$$

Now, let $x=0$ and $z=-y$. This tells us that

$$
f(y)^{3}+f(-y)^{3}=0 \Leftrightarrow f(-y)=-f(y), \forall y \in \mathbb{R}
$$

so $f$ is odd.

1 point.

Moreover, for $z=0$ and $y=-x$, we have

$$
f\left(x^{3}\right)+f(-x)^{3}=0 \Leftrightarrow f\left(x^{3}\right)=f(x)^{3}, \forall x \in \mathbb{R}
$$

because $f$ is odd.

1 point.

This means that the original equation becomes

$$
f(x)^{3}+f(y)^{3}+f(z)^{3}=3 x y z
$$

for all $x, y, z$ such that $x+y+z=0$, and by substituting $z=-x-y$ we can transform this into the equation (using also the fact that $f$ is odd),

$$
f(x)^{3}+f(y)^{3}+f(-x-y)^{3}=-3 x y(x+y) \Leftrightarrow f(x+y)^{3}=f(x)^{3}+f(y)^{3}+3 x y(x+y), \forall x, y \in \mathbb{R}
$$

Lemma: For all $n \in \mathbb{N}$ and $x \in \mathbb{R}$, the following relation holds:

$$
f(n x)^{3}=n f(x)^{3}+\left(n^{3}-n\right) x^{3}
$$

Proof. We proceed by induction on $n$. For $n=1$ the statement is obvious for all $x$. Next, if this relation holds for some $n \in \mathbb{N}$ and all $x$, then by putting $y \rightarrow n x$ in (1), we conclude that

$$
\begin{gathered}
f((n+1) x)^{3}=f(x)^{3}+f(n x)^{3}+3 n(n+1) x^{3}=(n+1) f(x)^{3}+\left(n^{3}-n+3 n(n+1)\right) x^{3} \\
(n+1) f(x)^{3}+\left((n+1)^{3}-(n+1)\right) x^{3}
\end{gathered}
$$

and the Lemma follows.

1 point.

Next, we fix $x \in \mathbb{R}$ and we compute $f\left(n^{3} x^{3}\right)^{3}=f(n x)^{9}$ in two different ways for all $n \in \mathbb{N}$. Of course,

$$
f\left(n^{3} x^{3}\right)^{3}=n^{3} f\left(x^{3}\right)^{3}+\left(n^{9}-n^{3}\right) x^{9}=n^{9} x^{9}+n^{3}\left(f(x)^{9}-x^{9}\right)
$$

On the other hand, we have

$$
\begin{aligned}
f(n x)^{9} & =\left(n f(x)^{3}+\left(n^{3}-n\right) x^{3}\right)^{3} \\
& =n^{3} f(x)^{9}+3 n^{2}\left(n^{3}-n\right) f(x)^{6} x^{3}+3 n\left(n^{3}-n\right)^{2} f(x)^{3} x^{6}+\left(n^{3}-n\right)^{3} x^{9} \\
& =n^{9} x^{9}+3 n^{7}\left(f(x)^{3} x^{6}-x^{9}\right)+3 n^{5}\left(f(x)^{6} x^{3}-2 f(x)^{3} x^{6}+x^{9}\right) \\
& +n^{3}\left(f(x)^{9}-3 f(x)^{6} x^{3}+3 f(x)^{3} x^{6}-x^{9}\right)
\end{aligned}
$$

3 points.

This means that the polynomials

$$
\begin{gathered}
p_{x}(y)=x^{9} y^{9}+\left(f(x)^{3}-x^{3}\right) y^{3} \quad \text { and } \\
q_{x}(y)=x^{9} y^{9}+3 x^{6}\left(f(x)^{3}-x^{3}\right) y^{7}+3 x^{3}\left(f(x)^{3}-x^{3}\right)^{2} y^{5}+\left(f(x)^{3}-x^{3}\right)^{3} y^{3}
\end{gathered}
$$

are equal for infinitely many values of $y$, thus they are equal as polynomials. Comparing the coefficients of $y^{7}$, we get that $f(x)^{3}=x^{3} \Leftrightarrow f(x)=x, \forall x \in \mathbb{R}$. This function obviously satisfies the condition of the problem.

4 points.