Quite tedious As always, let $P(x, y)$ denote the original assertion when $x, y \in \mathbb{Z}$. Also note that throughout the entire solution, when we use modulo some number $z$, we're also checking that $z \neq 0$. $P(x, x): f(2x)f(x^2) = 2f(x^3)$ meaning that $$f(2)f(1) = 2f(1), \,\, f(-2)f(1) = 2f(-1), \,\, f(0)^2 = 2f(0)$$Let's denote $(\star)$ to these sets of equations. From $(\star)$ we get $f(1) = 0$ or $f(2) = 2$ as well as $f(0) = 0$ or $f(0) = 2$. Case 1: $f(1) = 0, f(0) = 0$ $$P(x, 0): f(x)^3 = f(x^3)$$$$P(x, 1): f(x+1)f(x)(f(x)+1) = f(x^3) = f(x)^3$$ If $f(u) = -1$ for some $u \in \mathbb{Z}$ then we get $-1 = (-1)^3 = f(u^3) = 0 \,\, (\Rightarrow \Leftarrow)$. Thus, using modulo $f(x)+1$ we get $f(x)+1 \,|\, f(x)$ leading to $f(x)+1 \,|\, 1$ which implies that $f(x) \in \{0, -2\}$ for all $x \in \mathbb{Z}$. If $f(u) = -2$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = -8 \,\, (\Rightarrow \Leftarrow)$. Solution 1: $f(x) = 0 \,\, \forall \,\, x \in \mathbb{Z}$. Case 2: $f(1) = 0, f(0) = 2$ $$P(x, 1): f(x+1)f(x)(f(x)+1) = f(x^3)$$$$P(x, 0): f(x)(f(x)^2 - 4f(x) + 6) = f(x^3) + 2$$ If $f(u) = 0$ for some $u \in \mathbb{Z}$ then we get $0 = f(u^3) = -2 \,\, (\Rightarrow \Leftarrow)$. Thus, using modulo $f(x)$ we get $f(x) \,|\, 2$. If $f(u) = -1$ for some $u \in \mathbb{Z}$ then we get $0 = f(u^3) = -13 \,\, (\Rightarrow \Leftarrow)$. Thus, using modulo $f(x)+1$ we get $f(x)+1 \,|\, 13$. Combining these two results, we get $f(x) = -2$ for all $x \in \mathbb{Z}$. Checking in the original assertion $(\Rightarrow \Leftarrow)$. Case 3: $f(1) \neq 0, f(2) = 2, f(0) = 0$ $$P(x, 0): f(x)^3 = f(x^3) \,\, \Rightarrow \,\, f(8) = 8 \text{and} f(1), f(-1) \in \{-1, 0, 1\}$$ $$P(-1, 2): f(1)((f(-1)-2)^2 + f(-2)) = f(-1) + 8$$ If $f(-1) = 0$, by $(\star)$ we get $f(-2) = 0$ and then $f(1) = 2 \,\, (\Rightarrow \Leftarrow)$. If $f(1) = -1$, we make two subcases with $(\star)$ $\bullet f(-2) = 2, f(-1) = -1$, then $-11 = 7 \,\, (\Rightarrow \Leftarrow)$. $\bullet f(-2) = -2, f(-1) = 1$, then $1 = 9 \,\, (\Rightarrow \Leftarrow)$. If $f(1) = 1$, we make two subcases with $(\star)$ $\bullet f(-2) = 2, f(-1) = 1$, then $3 = 9 \,\, (\Rightarrow \Leftarrow)$. $\bullet f(-2) = -2, f(-1) = -1$, then $$P(x, 1): f(x+1)(f(x)^2 - f(x) + 1) = f(x^3) + 1 = f(x)^3 + 1 = (f(x)+1)(f(x)^2 - f(x) + 1)$$ Since $f(x)^2 - f(x) + 1 \neq 0$ we're left with $f(x+1) = f(x)+1$ for all $x \in \mathbb{Z}$. Solution 2: $f(x) = x \,\, \forall \,\, x \in \mathbb{Z}$. Case 4: $f(1) \neq 0, f(2) = 2, f(0) = 2$ $$P(x, 0): f(x)(f(x)^2 - 4f(x) + 6) = f(x^3) + 2$$ Using $P(1, 0)$ we get $f(1) \in \{1, 2\}$. Also $P(1, -1): 2(f(1) - f(-1))^2 = f(1) - f(-1)$ which means that $f(1) = f(-1)$. Case 4.1: $f(1) = f(-1) = 2$ $$P(x, 1): f(x+1)(f(x)^2 - 3f(x) + 4) = f(x^3) + 2$$ Let's denote temporarily $a := f(x)$. We have $a^2 - 3a + 4 \,|\, a(a^2 - 4a + 6)$, reducing it we end up with $a^2 - 3a + 4 \,|\, 4-a$. Also remember that $a^2 - 3a + 4 \neq 0$. Thus, by the inequalities between the absolute values of both sides, we get $a \in \{0, 2, 4\}$. If $f(u) = 4$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = 22, f(u+1) = 24/8 = 3 \,\, (\Rightarrow \Leftarrow)$. If $f(u) = 2$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = 2, f(u+1) = 2$ implying that $$\text{If } f(k) = 2, \text{ then } f(\geq k) = 2$$. If $f(u) = 0$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = -2, f(u+1) = 0$ implying that $$\text{If } f(k) = 0, \text{ then } f(\geq k) = 0$$. Since $f(0) = 2$ already, then $f(x) = 2 \,\, \forall \,\, x \geq 0$. If at any point $f(y) = 0$ for some $y$, then we would have $f(\text{sufficiently large non-negative integer}) = 0 \,\, (\Rightarrow \Leftarrow)$. Solution 3: $f(x) = 2 \,\, \forall \,\, x \in \mathbb{Z}$. Case 4.2: $f(1) = f(-1) = 1$ $$P(x, 1): f(x+1)(f(x)^2 - f(x) + 1) = f(x^3) + 1$$ Let's denote temporarily $a := f(x)$. We have $a^2 - a + 1 \,|\, a(a^2 - 4a + 6) - 1$, reducing it we end up with $a^2 - a + 1 \,|\, 2a + 2$. Also remember that $a^2 - a + 1 \neq 0$. Since $a^2 - a$ is even, then $a^2 - a + 1 \,|\, a+1$. Thus, by the inequalities between the absolute values of both sides, we get $a \in \{-1, 0, 1, 2\}$. If $f(u) = 0$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = -2 \,\, (\Rightarrow \Leftarrow)$. If $f(u) = -1$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = -13 \,\, (\Rightarrow \Leftarrow)$. If $f(u) = 1$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = 1, f(u+1) = 2$. If $f(u) = 2$ for some $u \in \mathbb{Z}$ then we get $f(u^3) = 2, f(u+1) = 1$. Solution 4: $f(x) = 2 - (x \mod 2) \,\, \forall \,\, x \in \mathbb{Z}$.

There are 4 answers, $$f(x)\equiv x, \qquad f(x)\equiv 0, \qquad f(x)\equiv 2, \qquad f(x)\equiv \begin{cases} 1 \quad \textrm{when $x$ is odd} \\ 2 \quad \textrm{when $x$ is even} \\ \end{cases} $$ It's easy to verify that all of the above solutions work. Let the functional equation be $P(x,y)$. $$P(0,0): \quad f(0)^2=2f(0) \Rightarrow f(0)=0 \textrm{ or } 2$$ We break them into 2 main cases: Case 1. $f(0)=0$. Then: $$P(x,0): \quad f(x)^3=f(x^3)$$ Thus we can rewrite $P(x,y)$ as $$f(x+y)((f(x)-f(y))^2+f(xy))=f(x)^3+f(y)^3$$ Now we have: $$P(1,1): \quad f(2)f(1)=2f(1) \Rightarrow f(1)=0 \textrm{ or } f(2)=2$$ We have 2 sub-cases: Case 1A. $f(1)=0$. Then: $$P(-1,-1): \quad f(-2)f(1)=2f(-1) \Rightarrow f(-1)=0$$ We now prove by induction that $f(x)\equiv 0$. The base case is $f(0)=0$. Now assume $f(k)=0$ for some integer $k$. Then: $$P(k,1): \quad f(k+1)(\textrm{something})=f(k)^3+f(1)^3=0 \Rightarrow f(k+1)=0$$$$P(k,-1): \quad f(k-1)(\textrm{something})=f(k)^3+f(-1)^3=0 \Rightarrow f(k-1)=0$$ Thus the induction is complete. Case 1B. $f(2)=2$. Then: $$P(1,0): \quad f(1)^3=f(1) \Rightarrow f(1)=-1,0 \textrm{ or }-1$$ Similarly, $f(-1)=-1,0 \textrm{ or }-1$ If $f(1)=0$, from Case 1A, we get $f(x)\equiv 0$, which contradicts $f(2)=2$. If $f(-1)=0$, we have: $$P(-1,-1): \quad f(-2)f(1)=2f(-1)=0 \Rightarrow f(-2)=0 \textrm{ or } f(1)=0$$ $f(1)=0$ leads to the contradiction above. $f(-2)=0$ means: $$P(2,-2): \quad f(0)(\textrm{something})=f(2)^3+f(-2)^3 \Rightarrow 0=8$$ This is a contradiction. Thus $f(1),f(-1)\neq 0$. Moreover, $$P(1,-1): \quad f(0)(\textrm{something})=f(1)^3+f(-1)^3 \Rightarrow f(1)+f(-1)=0$$ Therefore $(f(1),f(-1))=(1,-1)$ or $(-1,1)$. However, if $f(1)=-1$, we get $$P(2,1): \quad f(3)((f(2)-f(1))^2+f(2))=f(2)^3+f(1)^3 \Rightarrow 11f(3)=8$$which cannot happen. Thus $f(1)=1$ and $f(-1)=-1$. We are now ready to again execute our induction to prove that $f(x)\equiv x$. Note that it is already true in the range $[-1,1]$, which is our base case. Now assume that $f(x)=x$ on $[-k,k]$ for some integer $k$. We have: $$P(k,1): \quad f(k+1)((f(k)-f(1))^2+f(k))=f(k)^3+f(1)^3$$$$\Rightarrow f(k+1)(k^2-k+1)=k^3+1 \Rightarrow f(k+1)=k+1$$and $$P(-k,-1): \quad f(-k-1)((f(-k)-f(-1))^2+f(k))=f(-k)^3+f(-1)^3$$$$\Rightarrow f(-k-1)(k^2-k+1)=-k^3-1 \Rightarrow f(-k-1)=-k-1$$ Thus the claim is also true on $[-(k+1),k+1]$, which completes our induction, and completes Case 1. Case 2. $f(0)=2$. Then: $$P(x,0): \quad f(x)((f(x)-2)^2+2)=f(x^3)+2$$ This means, $f(x)=1 \Rightarrow f(x^3)=1$ and $f(x)=2 \Rightarrow f(x^3)=2$. Moreover, $$P(1,0): \quad f(1)((f(1)-2)^2+2)=f(1)+2 \Rightarrow (f(1)-2)(f(1)-1)^2=0$$ Thus $f(1)=1$ or 2. Similarly, using $P(-1,0)$, we get $f(-1)=1$ or 2. Moreover, $$P(1,-1): \quad f(0)(\textrm{something})=f(1)+f(-1) \Rightarrow f(1)+f(-1) \textrm{ is even}$$ Therefore $f(1)=f(-1)=1$ or $f(1)=f(-1)=2$. We again have 2 cases: Case 2A. $f(1)=f(-1)=1$. From $P(x,0)$ above, $f(x)=1 \Rightarrow f(x^3)=1$ and $f(x)=2 \Rightarrow f(x^3)=2$. We will prove by, once again, induction, that $$f(x)\equiv \begin{cases} 1 \quad \textrm{when $x$ is odd} \\ 2 \quad \textrm{when $x$ is even} \\ \end{cases}$$ Since $f(1)=f(-1)=1$ and $f(0)=2$, the claim is true on $[-1,1]$. Now assume that it's also true on $[-k,k]$ for some integer $k$. If $k$ is odd, $f(k)=f(-k)=1$ and thus $f(k^3)=f(-k^3)=1$. Now: $$P(k,1): \quad f(k+1)((f(k)-f(1))^2+f(k))=f(k^3)+f(1)$$$$\Rightarrow f(k+1)((1-1)^2+1)=1+1 \Rightarrow f(k+1)=2$$ and $$P(-k,-1): \quad f(-k-1)((f(-k)-f(-1))^2+f(k))=f(-k^3)+f(-1)$$$$\Rightarrow f(-k-1)((1-1)^2+1)=1+1 \Rightarrow f(-k-1)=2$$ If $k$ is even, similarly, using $P(k,1)$ and $P(-k,-1)$, we get $f(k+1)=f(-k-1)=1$. Thus the claim is also true on $[-(k+1),k+1]$ and the induction is complete. Case 2B. $f(1)=f(-1)=2$ We, for the final time, prove by induction that $f(x)\equiv 2$. Since $f(-1)=f(0)=f(1)=2$, it is true on $[-1,1]$. Now assume that $f(x)=2$ on $[-k,k]$ for some integer $k$. Then: $$P(k,1): \quad f(k+1)((f(k)-f(1))^2+f(k))=f(k^3)+f(1)$$$$\Rightarrow f(k+1)((2-2)^2+2)=2+2 \Rightarrow f(k+1)=2$$ and $$P(-k,-1): \quad f(-k-1)((f(-k)-f(-1))^2+f(-k))=f(-k^3)+f(-1)$$$$\Rightarrow f(-k-1)((2-2)^2+2)=2+2 \Rightarrow f(-k-1)=2$$ Therefore, the claim is also true on $[-(k+1),k+1]$. Thus the induction is complete and the problem is finally solved.

There are $4$ solutions work, which is $f\equiv x$, $f=\begin{cases} \ 1\ \text{if} \ x\equiv 1\pmod 2 \\ \ 2 \ \text{if} \ x\equiv 0\pmod {2} \end{cases}, \ f(x)\equiv 2, \ f(x)\equiv 0.$ Let $P(x,y)$ denote the origianl assertion : $f(x+y)((f(x) - f(y))^2+f(xy))=f(x^3)+f(y^3), \quad\forall x,y\in\mathbb{Z}$. $P(x,x)\implies f(2x)f(x^2)=2f(x^3)\quad\forall x\in\mathbb{Z}$. Let $x=0$ we have $f(0)=0$ or $f(0)=2$. $\textbf{Case 1:} \ f(0)=2.$ $P(x,0)\implies f(x)(f^2(x)-4f(x)+6)=f(x^3)+2\quad\forall x\in\mathbb{Z}$. Let $x=1$ and $f(1)=c$ then $(c-2)(c-1)^2=0$, meaning that $f(1)=1$ or $f(1)=2$. $\textbf{Case 1.1:} \ f(1)=2.$ Then $P(x,1): f(x^3)+2=f(x+1)(f^2(x)-3f(x)+4)=f(x)(f^2(x)-4f(x)+6)\quad\forall x\in\mathbb{Z}. \ (1)$ That is $$f(x+1)=\frac{f(x)(f^2(x)-4f(x)+6)}{f^2(x)-3f(x)+4}\quad\forall x\in\mathbb{Z}.$$By easy induction we get that $f(x)=2\quad\forall x\in\mathbb{N}\cup \{0\}.$ If we have $f(x)=2$ for some $x$, let $a=f(x-1)$ then $\frac{a^3-4a^2+6a}{a^2-3a+4}=2\iff (a-2)^3=0\iff a=2$. Hence,in this case, the first solution is $$\boxed{f(x)=2\quad\forall x\in\mathbb{Z}}$$. $\textbf{Case 1.2:}\ f(1)=1.$ $$P(x,1)\implies f(x+1)(f^2(x)-f(x)+1)=f(x^3)+1=f(x^3)+2-1=f(x)(f^2(x)-4f(x)+6)-1\quad\forall x\in\mathbb{Z}.$$That is $$f(x+1)=\frac{f^3(x)-4f^2(x)+6f(x)-1}{f^2(x)-f(x)+1}\quad\forall x\in\mathbb{Z}.\ (2)$$If $f(x)=2$ for some $x$ then we have $f(x+1)=1$, similarly if $f(x)=1$ then $f(x+1)=2$. If $f(x)=2$ then let $x\to x-1$ in $(2)$ we have $$(f(x-1)-1)(f^2(x-1)-5f(x-1)+3)=0\implies f(x-1)=1.$$If $f(x)=1$ then let $x\to x-1$ in $(2)$ we have that $$(f(x-1)-2)(f^2(x-1)-3f(x-1)+1)=0\implies f(x-1)=2.$$We also have $f(0)=2$ and $f(1)=1$, by easy induction we also claim the second solution is $$\boxed{f=\begin{cases} \ 1\ \text{if} \ x\equiv 1\pmod 2 \\ \ 2 \ \text{if} \ x\equiv 0\pmod {2}. \end{cases}}$$$\textbf{Case 2:} \ f(0)=0.$ $P(x,0)\implies f^3(x)=f(x^3)\quad\forall x\in\mathbb{Z}.$ Let $x=1$ then $f(1)=\pm 1$ or $0$. $\textbf{Case 2.1:} \ f(1)=0.$, then $P(x,1)$ we have $$f(x+1)(f^2(x)+f(x))=f(x^3)=f(x)^3\quad\forall x\in\mathbb{Z}$$If $f(x)=0$ then let $x\to x-1$ implies that $f(x-1)=0$, we also have $f(0)=0$ then $f(x)=0\quad\forall x\in\mathbb{Z}, x\le 0 \ (3)$. $$P(x,-x)\implies f(x^3)+f(-x^3)=0\implies f^3(x)+f^3(-x)=0\quad\forall x\in\mathbb{Z}.$$By $(3)$ we get the third solution is $$\boxed{f(x)\equiv 0\quad\forall x\in\mathbb{Z}}.$$$\textbf{Case 2.2:}\ f(1)=1.$ $$P(x,1)\implies f(x+1)=\frac{f(x^3)+1}{f^2(x)-f(x)+1}=\frac{f^3(x)+1}{f^2(x)-f(x)+1}=f(x)+1\quad\forall x\in\mathbb{Z}.$$Since $f(0)=0, f(1)=1$ and $f^3(x)=-f^3(-x)$, we claim the fourth sol is $$\boxed{f(x)=x\quad\forall x\in\mathbb{Z}}.$$$\textbf{Case 2.3:}\ f(1)=-1$. $P(1,1)\implies f(2)=2$. $$P(2,1)\implies f(3)(9+2)=f(2^3)+f(1)=f^3(2)+f(1)=7\implies f(3)\notin \mathbb{Z} \ (\text{contracdition}).$$So there are not solution in this case, findnally we have $4$ solutions metioned above work.

This problem was proposed by me, along with Q2. Solution is attached below.

Attachments:
My Problem 5.pdf (106kb)