The only solutions are $f(x)=x,-x$ the zero function and $\pm f(x)= \begin{cases} 0 & \text{if } x \equiv 0 \pmod 3 \\ 1 & \text{if } x \equiv 1 \pmod 3 \\ -1 & \text{if } x \equiv 2 \pmod 3 \end{cases}$.

Let $P(x,y,z)$ denote the assertion. Clearly $f\equiv 0$ is a solution, so assume $f$ is non-constant. $P(0,0,0)\Longleftrightarrow f(0)=0$ $P(x,0,0)\Longleftrightarrow f(x)^3=f(x^3) \Rightarrow f(1)=\pm1$(why?If $f(1)=0$ then $P(x,1,0)$ implies that f is constant.) $P(x,-x,0)\Longleftrightarrow f(x)=-f(-x)$ If $f(1)=1$ take $P(x,1,0)$ and then induct to show that $f(x)=x$ or $f(x) = \begin{cases} 0 & \text{if } x \equiv 0 \pmod 3 \\ 1 & \text{if } x \equiv 1 \pmod 3 \\ -1 & \text{if } x \equiv 2 \pmod 3 \end{cases}$ If $f(1)=-1$ take $P(x,-1,0)$ and induct again to show $f(x)=-x$ or $f(x) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ -1 & \text{if } x \equiv 1 \pmod 3 \\ 1 & \text{if } x \equiv 2 \pmod 3 \end{cases}$

The only solutions are $f(x)=x,-x$ and the zero function.

Let $P(x,y,z)$ denote the assertion. Clearly $f\equiv 0$ is a solution, so assume $f$ is non-constant. $P(0,0,0)\Longleftrightarrow f(0)=0$ $P(x,0,0)\Longleftrightarrow f(x)^3=f(x^3) \Rightarrow f(1)=\pm1$ $P(x,-x,0)\Longleftrightarrow f(x)=-f(-x)$ If $f(1)=1$ take $P(x,1,0)$ and then induct to show that $f(x)=x$ If $f(1)=-1$ take $P(x,-1,0)$ and induct again to show $f(x)=-x$

The only solutions are $f(n)=0,n,-n,$ $$f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ 1 & \text{if } n \equiv 1 \pmod 3 \\ -1 & \text{if } n \equiv 2 \pmod 3 \end{cases}\text{ and }f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ -1 & \text{if } n \equiv 1 \pmod 3 \\ 1 & \text{if } n \equiv 2 \pmod 3 \end{cases}$$

Let $P(a,b,c)$ denote the assertion $ f(a^3) + f(b^3) + f(c^3) + 3f(a + b)f(b + c)f(c + a) = {(f(a + b + c))}^3$. Firstly, $P(0,0,0)$ clearly implies $f(0)=0$ which implies, from $P(a,0,0)$, $f(a^3)=f(a)^3$ (a fact that we will commonly use below). Then $P(1,0,0)\implies f(1)=f(1)^3$, so $f(1)$ is either $0,-1,$ or $1$. Case 1: $f(1)=0$. We claim that by induction, $f(n)=0$ for all $n\geq 0$. Skipping the obvious base case, suppose $f(k)=0$. Then $P(k,1,0)\implies 0=f(k)^3=f(k+1)^3 $ hence $f\equiv 0$ over the positive integers. Also, $P(1,-1,0)\implies f(-1)=0$ - which will be the base case for our inductive proof that $f(n)=0$ for all $n\leq 0$. Indeed, suppose $f(-k)=0,k>0.$ Then $P(-k,-1,0)\implies f(-k-1)=0$ completing the inductive proof, and hence $f(n)=0$ for all $n\in\mathbb{Z}$. Case 2: $f(1)=1$. $P(1,1,0)\implies 2+3f(2)=f(2)^3\implies (f(2)-2)(f(2)+1)^2=0$ hence $f(2)=-1$ or $2$. Subcase 2.1: $f(2)=-1$. Then $P(1,1,1)\implies f(3)=0$, and it can be easily derived that $f(4)=1$, $f(5)=-1$, and $f(6)=0$. In fact, this pattern continues: with these being our base cases, we claim by induction that $ f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ 1 & \text{if } n \equiv 1 \pmod 3 \\ -1 & \text{if } n \equiv 2 \pmod 3 \end{cases}$ To prove this, suppose this is true for $f(k)$. Then consider $f(k+1)$. If $k+1\equiv 0\pmod 3$ ,then $P(k-2,3,0)\implies f(k+1)=0$. If $k+1\equiv 1\pmod 3$, then $P(k,1,0)\implies f(k+1)^3=1\implies f(k+1)=1$ since $f(k)=0$ as $3|k$. If $k+1\equiv 2\pmod 3$, then $P(k-1, 2, 0) \implies f(k+1)^3=-1\implies f(k+1)=-1$ since $f(k-1)=0$, as $3|k-1$. This completes the induction. Subcase 2.2: $f(2)=2$. Note that $P(1,1,1)\implies f(3)=3$. With these as our base cases, we claim that $f(n)=n$ for all $n\geq 0$ by induction. Suppose $f(k)=k$. Then $P(k,1,0)\implies k^3+1+3kf(k+1)=f(k+1)^3$. Let $x=f(k+1)$, so that this becomes $x^3-3kx-k^3-1=0$, or $$(x^2+x+1+k^2+kx-k)(x-k-1)=0$$which implies either $x^2+x+1+k^2+kx-k$ or $x=k+1$. But the former cannot happen - treating the term as a quadratic in $x$, its discriminant is $(k+1)^2-4(k^2-k+1)=-3(k+1)^2$ which is clearly always negative. Hence $x=f(k+1)=k+1$. To extend this, $P(-1,1,0)\implies f(-1)=-1$ - our base case. To induct, suppose $f(-k)=-k,k>0.$ Then $P(-k,-1,0)\implies x^3-3kx+k^3+1=(k + x + 1) (k^2 - k x - k + x^2 - x + 1)=0$, where $x=f(-k-1)$. Notice that $k^2-kx-k+x^2-x+1=x^2-(k+1)x+k^2-k+1$, and if we see this as a quadratic in $x$, its discriminant is the same as quadratic discussed in Subcase 2.2. Hence $k+x+1=0$, or $x=f(-k-1)=-k-1$ as desired, which means $f(n)=n$ for all $n\in \mathbb{Z}$. Case 3: $f(1)=-1$ $P(1,1,0)\implies f(2)^3-3f(2)+2=(f(2)-1)^2(f(2)+2)=0$ so $f(2)=1$ or $-2$. Subcase 3.1: $f(2)=1$. $P(1,1,1)\implies f(3)=0$. We can again determine $f(4)=-1, f(5)=1, f(6)=0$ and then it's easy to inductively prove $$f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ -1 & \text{if } n \equiv 1 \pmod 3 \\ 1 & \text{if } n \equiv 2 \pmod 3 \end{cases}$$as in Subcase 2.1. Subcase 3.2: $f(2)=-2$. Then $P(1,1,1)\implies f(3)=-3$. one can easily do the same as in Subcase 2.2 to find $f(n)=-n$ for all $n\in \mathbb{Z}$. Concluding, the only solutions are $f(n)=0,n,-n,$ $$f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ 1 & \text{if } n \equiv 1 \pmod 3 \\ -1 & \text{if } n \equiv 2 \pmod 3 \end{cases}\text{ and }f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod 3 \\ -1 & \text{if } n \equiv 1 \pmod 3 \\ 1 & \text{if } n \equiv 2 \pmod 3 \end{cases}$$I invite the reader to check that these functions work.

$f(x) = x$, $f(x) = -x$, and $f(x) = 0$.

Plugging in $x = y = z = 0$ gives us $3f(0) + 2f(0)^3 = 0$, so $f(0) = 0$. Next, plugging in $y = z = 0$ gives us: $$ f(x^3) = f(x)^3 \; \forall x\in \mathbb{R} \quad \ldots(1) $$ The problem becomes: $$ f(x)^3 + f(y)^3 + f(z)^3 + 3f(x+y)f(y+z)f(z+x) = f(x+y+z)^3 \; \forall x, y, z\in \mathbb{R} \quad \ldots(2) $$ First, we will just solve $(2)$. Functions that satisfy (2)All additive functions $f$ over $\mathbb{R} \rightarrow \mathbb{R}$. ProofIt is easy to check that all additive functions $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfy $(2)$. Now, to prove that it is additive, we observe again that plugging in $x = y = z = 0$ gives us $6f(0)^3 = f(0)^3$ and thus $f(0) = 0$. Then, plugging in $y = -x$ and $z = 0$, we obtain that $f$ is odd. Now, plugging in $z = -x-y$ (so that $x+y+z = 0$), we get $$ f(x)^3 + f(y)^3 + f(z)^3 = 3f(x)f(y)f(z) $$ Thus, for every real numbers $x$ and $y$: \begin{align*} f(x) = f(y) = f(-x-y) \quad &\lor \quad f(x)+f(y)+f(-x-y) = 0 \\ f(x+y) = -f(x) = -f(y) \quad &\lor \quad f(x+y) = f(x) + f(y) \quad \ldots(3) \end{align*} To prove that $f(x+y) = f(x) + f(y)$ for all $x, y\in \mathbb{R}$, we will prove a claim. ClaimIf $f(x) \neq 0$, then $3 f\left(\frac{x}{3}\right) = f(x)$. ProofFrom $(3)$, we know that either $f\left(\frac{2x}{3}\right) = -f\left(\frac{x}{3}\right)$ or $f\left(\frac{2x}{3}\right) = 2f\left(\frac{x}{3}\right)$. Either way, $f(x) = -f\left(\frac{2x}{3}\right) = -f\left(\frac{x}{3}\right)$ would yield $f(x) = 0$; a contradiction. So, by $(3)$, $f(x) = f\left(\frac{2x}{3}\right) + 2f\left(\frac{x}{3}\right)$. The former case again yields $f(x) = 0$, while the latter yields $f(x) = 3f\left(\frac{x}{3}\right)$, as desired. Now, suppose for the sake of contradiction that $f(x+y) \neq f(x) + f(y)$ for some real number $x$ and $y$. Then, $f(x+y) = -f(x) = -f(y) \neq 0$. By Claim, $f(y) = f(x) = 3 f\left(\frac{x}{3}\right)$. Hence, applying $(3)$ several times (with former case being impossible), \begin{align*} f\left(\frac{x}{3} + y\right) &= f\left(\frac{x}{3}\right) + f(y) \\ &=\frac{4}{3} f(x) \\ f\left(\frac{2x}{3} + y\right) &= f\left(\frac{x}{3}\right) + f\left(y + \frac{x}{3}\right) \\ &= \frac{5}{3} f(x) \\ f(x+y) &= f\left(\frac{x}{3}\right) + f\left(y + \frac{2x}{3}\right) \\ &= 2f(x) \\ -f(x) &= 2f(x) \\ f(x) &= 0 \end{align*} A contradiction. Hence, indeed $f$ is additive. Rest of the solutionNow, we prove that if $f$ is additive and satisfy $(1)$, then $f$ must be linear. This then yields the answer above. Since $f$ is additive, for every positive integer $k$ and real number $x$, \begin{align*} f((x+k)^3) &= f(x+k)^3 \\ f(x^3) + 3kf(x^2) + 3k^2 f(x) + k^3 f(1) &= f(x)^3 + 3kf(x)^2 f(1) + 3k^2 f(x) f(1)^2 + k^3 f(1)^3 \end{align*} Treating the equation as polynomial equation in terms of $k$, we get $f(x)^2 f(1) = f(x^2)$ for all $x\in \mathbb{R}$. In particular, the image $f(\mathbb{R}_{\geq 0})$ is either bounded above (if $f(1) \leq 0$), or bounded below (if $f(1) > 0$). Either way, $f$ must be linear. Also, it is easy to check by comparing again that $f(1) \in \{-1, 0, 1\}$, so we are done.