@above, that $f$ doesn't map into $\mathbb{Z}$...

... which proves that no such function exists.

Don't forget that $f(x)\equiv \frac{x^3-x}{3}+tx$ are also solutions for any $t\in\mathbb{Z}.$ These can be proved to be only solutions by induction. Kaskade wrote:

Unfortunately, the last four equation are dependent, so you cannot conclude that $f(x)=\frac{x^3}{3}.$

This problem relies on the identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.

$x=y=z=0 \ \implies \ f(0)=0$ $y=-x, z=0 \ \implies \ f(-x)=-f(x)$ $y=-x-1, z=1 \ \implies \ f(x+1)-f(x)=x^2+x+f(1)$ so, inducting, $f(x)=cx+\frac{x^3-x}{3},$ which works for all $c=f(1)$

Let $P(x,y,z)$ be the assertion. $P(0,0,0): 3f(0)=0 \Rightarrow f(0)=0$ $P(x,y,-x-y): f(x)+f(y)+f(-x-y)=-xy(x+y)$ $P(0,x+y,-x-y): f(0)+f(x+y)+f(-x-y)=0$ $\Rightarrow f(x+y)-f(x)-f(y)=xy(x+y) \Rightarrow f(x+y)=f(x)+f(y)+xy(x+y)$ $\mathit (i)$ Now we define $g(x)=f(x) - \frac{x^3}{3}$ So, according to $ \mathit (i), we have g(x+y)=g(x)+g(y)$ and we have $g: \Bbb{Q} \rightarrow \Bbb{Q}$ so $g(x)=cx \Rightarrow f(x)=cx + \frac{x^3}{3}$ Which is a valid solution.

Note that $\frac{x^3}3+\frac{y^3}3+\frac{z^3}3=xyz$, so if $g:\mathbb Z\to\frac13\mathbb Z$ is defined by $g(x)=f(x)-\frac{x^3}3$ we derive that $g(x)+g(y)+g(z)=0$. Let this assertion be $P(x,y,z)$ for $x,y,z$ such that $x+y+z=0$. Then $P(0,0,0)$ implies that $g(0)=0$, and $P(x,-x,0)$ implies that $g(-x)=-g(x)$. Finally we use $P(x,y,-x-y)$ in combination with the fact that $g$ is odd to get $g(x+y)=g(x)+g(y)$. Now by Cauchy, $g(x)=cx$ for some constant $c\in\frac13\mathbb Z$. Then, $f(x)=\frac{x^3}3+cx$. However, $f(x)$ is always an integer, so we require $f(x)=\frac{x^3-x}3+cx$ for some constant $c\in\mathbb Z$, which indeed satisfy the equation.

Clearly $f(0)=0.$ And setting $y=-x$ and $z=0$ $\implies$ $f$ is odd. And so setting $z=-x-1$ and $y=1$ implies $f(x+1)-f(x)=x+x^2+f(1)$ and by induction, $f(x)=(x^3-x)/3+xf(1),$ which works.

Good Problem! My sol. is sim. to #9, but anyway: $P(0,0,0): f(0)=0$ $P(x,-x,0): f(-x)=-f(x)$ $P(x+y,-x,-y): xy(x+y) = f(x+y)+f(-x)+f(-y) = f(x+y)-f(x)-f(y) = ((x+y)^3-x^3-y^3)/3$ $(*)$. Then let $g(x)=f(x)-x^3/3$, then from $(*)$ we have $g(x+y)=g(x)+g(y)$, since we solve for integers $x,y$, we can obtain by Cauchy that $g(x)=cx$ and since $c=f(1)-1/3=k-1/3$ where k is a positive integer, we can find that $f(x)=(x^3-x)/3+kx$, which satisfies the given condition. Then answer is $f(x)=(x^3-x)/3+kx$, where $k$ is an any positive integer

X.Allaberdiyev wrote: Good Problem! My sol. is sim. to #9, but anyway: $P(0,0,0): f(0)=0$ $P(x,-x,0): f(-x)=-f(x)$ $P(x+y,-x,-y): xy(x+y) = f(x+y)+f(-x)+f(-y) = f(x+y)-f(x)-f(y) = ((x+y)^3-x^3-y^3)/3$ $(*)$. Then let $g(x)=f(x)-x^3/3$, then from $(*)$ we have $g(x+y)=g(x)+g(y)$, since we solve for integers $x,y$, we can obtain by Cauchy that $g(x)=cx$ and since $c=f(1)-1/3=k-1/3$ where k is a positive integer, we can find that $f(x)=(x^3-x)/3+kx$, which satisfies the given condition. Then answer is $f(x)=(x^3-x)/3+kx$, where $k$ is an any positive integer I would be cautious. Your $g$ is a function $g:\mathbb Z \to \mathbb Z/3$ and assuming that Cauchy = Linear for this class of functions might need some convincing. Now, in this case you could instead define $g(x) = 3f(x) - x^3$ and everything is happy once more.