Start by plugging in numbers that satisfy $x+y+z=0$ $(0, 0, 0) \rightarrow f(0) + f(0) + f(0) = 0$ thus $f(0)=0$ $(0, x, -x) \rightarrow f(0) + f(x)^3 + f(-x)^3 = 0 \rightarrow f(x)^3 = -f(-x)^3$ meaning $f(-x)=-f(x)$ (f is odd) $(x, 0, -x) \rightarrow f(x^3) + f(-x)^3 = 0$ but since f is odd, $\rightarrow f(x^3) + (-f(x))^3 = 0$ thus $f(x^3) = f(x)^3$ Where to go from here?

physicskiddo wrote: Start by plugging in numbers that satisfy $x+y+z=0$ $(0, 0, 0) \rightarrow f(0) + f(0) + f(0) = 0$ thus $f(0)=0$ $(0, x, -x) \rightarrow f(0) + f(x)^3 + f(-x)^3 = 0 \rightarrow f(x)^3 = -f(-x)^3$ meaning $f(-x)=-f(x)$ (f is odd) $(x, 0, -x) \rightarrow f(x^3) + f(-x)^3 = 0$ but since f is odd, $\rightarrow f(x^3) + (-f(x))^3 = 0$ thus $f(x^3) = f(x)^3$ Where to go from here? Correction: $(0,0,0)$ we have $f(0)^3+2f(0)=0 \Rightarrow f(0)=0$

Here is a troll solution by one of the Bulgarian contestants. (Hard to expect that something like this would work in P3$\ldots$) Note that $f(x) = x$ works simply because of the identity $x^3 + y^3 + z^3 = 3xyz$ for $x+y+z = 0$. (In general, their difference is $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.) We claim it is the only one. Setting $x=y=z=0$ gives $f(0)=0$, then $x=0$, $y=-z$ gives that $f$ is odd. Then $y=-x$ and $z=0$ gives $f(x^3) = f(x)^3$. So we move on to $f(x)^3 + f(y)^3 + f(z)^3 = 3xyz$ with $f(x^3) = f(x)^3$. Setting $y=x$, $z=-2x$ gives $2f(x^3) - f(8x^3) = -6x^3$. We can replace $x^3$ by $x$ everywhere to get $2f(x) - f(8x) = -6x$. Now we need to express $f(8x)$ in another way. Setting $y=x$, $z=-2x$ also gives $2f(x)^3 - f(2x)^3 = -6x^3$, hence $f(2x) = \sqrt[3]{2f(x)^3 + 6x^3}$. Replacing $x$ with $2x$ now gives $f(4x) = \sqrt[3]{2f(2x)^3 + 48x^3} = \sqrt[3]{4f(x)^3 + 60x^3}$ and doing it once again leads to $f(8x) = \sqrt[3]{2f(4x)^3 + 384x^3} = \sqrt[3]{8f(x)^3 + 504x^3}$. Therefore $2f(x) + 6x = f(8x) = \sqrt[3]{8f(x)^3 + 504x^3}$, i.e. $f(x) + 3x = \sqrt[3]{f(x)^3 + 63x^3}$. This is equivalent to $9x(f(x)+4x)(f(x) - x) = 0$. So for all $x$ (including $0$, as $f(0) = 0$) we have $f(x) = x$ or $f(x) = -4x$. If we suppose that $f(x_0) = -4x_0$ for some $x_0$, then setting $x=y=x_0$, $z=-2x_0$ gives $-128x_0^3 - f(2x_0)^3 = -6x_0^3$, i.e. $f(2x_0)^3 = -122x_0^3$ which is impossible no matter whether $f(2x_0)$ is $2x_0$ or $-8x_0$. Therefore $f(x) = x$ for all $x$, as desired.

Let $P(x,y,z)$ denote the assertion of the given functional equation. Note that $P(0,0,0)$ gives us that $f(0)+f(0)^3+f(0)^3=0 \implies f(0)(1+2f(0)^2)=0 \implies f(0)=0$. Moreover, $P(0,x,-x)$ yielfs: $$ f(x)^3+f(-x)^3=0 \implies f(-x)=-f(x) $$Now we consider $P(x,-x,0)$: $$ f(x^3)+f(-x)^3 =0 \implies f(x^3) = f(x)^3 $$This allows us to rewrite the original equation as follows : $$ f(x)^3+f(y)^3+f(z)^3 = 3xyz $$for all real numbers $x,y,z$ satisfying $x+y+z=0$. It is easy to prove by mathematical induction that for every positive integer $n$ we have $f(nx)^3 =nf(x)^3 + (n^3-n)x^3$. Consider this expression for $n=8$: $$ f(8x)^3 = 8f(x)^3 +504x^3 $$Suppose that $x=a^3$ for some real number $a$. \begin{align*} f(8a^3)^3 = 8f(a^3)^3 + 504a^9 \\ (f(2a)^3)^3 =8(f(a)^9 + 504a^9 \\ (2f(a)^3+6a^3)^3 =8(f(a)^9+63a^9) \\ 8(f(a)^3+3a^3)^3 =8(f(a)^9+63a^9 ) \\ (f(a)^3+3a^3)^3 = (f(a)^9+63a^9) \\ f(a)^9+9f(a)^6a^3+27f(a)^3a^6+27a^9 =f(a)^9+63a^9 \\ f(a)^6a^3+3f(a)^3a^6-4a^9=0 \\ f(a^3)^2 +3f(a^3)a^3 -4(a^3)^2 =0 \\ f(x)^2+3f(x)x -4x^2 =0 \\ (f(x)-x)(f(x)+4x) =0 \\ \end{align*}This means that for all $x$ we have $f(x)=x$ or $f(x)=-4x$. Suppose there is real number $x_0 \neq 0$ such that $f(x_0)=-4x_0$, then: $$ f(x_0)^3 = -64x_0^3 = f(x_0^3) $$But we know that $f(x_0^3) =x_0^3$ or $f(x_0)=-4x_0^3$. In both cases this yields $x_0=0$, so we have derived a contradiction. This means that $f(x)=x$ is for all real numbers $x$. It is easy to check that this function indeed works

Here is my solution. Let $P(x,y,z)$ be the given functional equation. $P(0,0,0): 2f(0)^{3}+f(0)=0$ $\iff f(0)=0$. $P(0,x,-x): f(x)=-f(-x)$, so $f$ is odd. $P(x,0,-x): f(x^{3})=f(x)^{3}$ Now if we look at $P(x,y,-x-y)$ for arbitrary $x,y$, we get that $f(x+y)^{3}-(x+y)^{3}=f(x)^{3}-x^{3}+f(y)^{3}-y^{3}, \forall x,y \in \mathbb{R}$ It means that the function $g(x)=f(x)^{3}-x^{3}=f(x^{3})-x^{3}$ is additive. Also we have $g(x^{3})=f(x^{3})^{3}-x^{9}$ - that gives us two expressions for $f(x^{3})$. From there we get $(g(x)+x^{3})^{3}=g(x^{3})+x^{9}$ $\iff g(x)^{3}+3g(x)^{2}x^{3}+3g(x)x^{6}=g(x^{3})$. From the additivity of $g$ we get that $g(nx)=ng(x)$ for every $n\in\mathbb{N}$. Now plugging $x=nx$ and dividing by $n^{7}$ gives us the following: $\frac{g(x)^{3}}{n^{4}}+\frac{3g(x)^{2}x^{3}}{n^{2}}+3g(x)x^{6}=\frac{g(x^{3})}{n^{4}}$. When $n$ tends to infinity we have $3g(x)x^{6}=0 \iff g(x)=0, \forall x\neq 0$. The last one means that $f(x)=x, \forall x \in \mathbb{R}$, which clearly satisfies the given functional equation.

@Strudan_Borisov An amazing way to make use of Cauchy's equation, congratulations!

Is there a clean solution? I got $f$ is odd, $f(0) = 0$, $f(x^3) = f(x)^3$ and I also got that $f({\sqrt{2}}^n) = \sqrt{2}^n \hspace{0.20cm} \forall n \in \mathbb{Z}$. Could this last observation help in any way?

Let $P(x,y,z)$ be the assertion in given equation. Define $g(x) = f(x)^3 - x^3$. We will show that $g \equiv 0$. $P(0,0,0) : f(0) = 0$. Then $P(x,0,-x)$ and $P(0,x,-x)$ together imply $f(x)^3 = f(x^3)$ and $f(-x) = -f(x)$. This also implies that $g(-x) = -g(x)$. Recollect the identity that for $x+y+z=0$, $x^3+y^3+z^3=3xyz$, . Hence, the given equation rearranges to for $x+y+z=0$, $g(x) + g(y) + g(z) = 0$. Hence, $g(x) + g(y) = -g(z) = -g(-x-y) = g(x+y)$ implying $g$ is additive. Hence $g(rx)$ = $rg(x)$ for all $r \in \mathbb{Q}$. Recollect $f(x)^3 = f(x)^3$. Hence $(g(x) + x^3)^3 = (f(x)^3)^3 = (f(x^3))^3 = g(x^3)+(x^3)^3$. This simplifies $g(x)^3 + 3g(x)x^3 (g(x) + x^3) = g(x^3)$. Replace $x$ by $rx$ (for $r \in \mathbb{Q}_{\neq 0,\pm 1}$ and using that $g(rx) = rg(x)$, we will get the new equation : $g(x)^3 + 3r^2g(x)x^3 (g(x) + r^2x^3) = g(x^3)$ . Subtract the two equations. If $g(x) \neq 0$, we simply get $g(x) = x^3(1+r^2)$ which is contradiction on varying $r$. Hence $g(x) = 0$ for all $x$, and hence $f(x) = x$ for all $x$. And we are done!