SELECTION EXAMINATION

Let $\mathrm{deg}P(x)=0$. Then $P(x)=a\ne 0$ and from relation (1) we have: $a^3+3a^2=-2a\iff a=-1$ or $a=-2$. Hence the constant (nonzero) polynomials $P(x)=-1$ and $P(x)=-2$, are solutions of the problem.

Let $\mathrm{deg}P(x)=n>0$. Then the polynomial $P(x)$ can be written as $$P(x)=a{x}^n+Q(x),\text{ with }a\ne 0\text{ and }\mathrm{deg}Q(x)=k\le n-1\text{ or }Q(x)=0.$$ By substitution into relation (1) we get: $$(a{x}^n+Q(x){)}^3+3(a{x}^n+Q(x){)}^2=a{x}^{3n}+Q(x^3)-3Q(-x)-3a\cdot (-1{)}^n{x}^n,(2)$$ for all $x\in \mathbb{R}$. Equating the coefficients of $x^{3n}$ in the two parts we find: $$a^3=a\iff a=1\text{ or }a=-1.$$ Therefore we distinguish the following cases:

I. $a=1$. Then relation (2) becomes: $$\begin{gathered} (x^n+Q(x){)}^3+3(x^n+Q(x){)}^2=x^{3n}+Q(x^3)-3Q(-x)-3(-1{)}^n{x}^n \\ \iff A(x)=B(x),(3) \end{gathered}$$ where we have put $$\begin{gathered} A(x)=3x^{2n}Q(x)+3x^n(Q(x){)}^2+(Q(x){)}^3+3x^{2n}+6x^{n}Q(x)+3(Q(x){)}^2 \\ B(x)=Q(x^3)-3Q(-x)-3(-1{)}^n{x}^n. \end{gathered}$$ If $Q(x)=0$, then: $3x^{2n}=-3\cdot (-1{)}^n{x}^n$, (impossible). If $\mathrm{deg}Q(x)=k>0$, then, since $0<k<n$, we have: $$2n+k=\mathrm{deg}A(x)=\mathrm{deg}B(x)=\max \{3k,n\}\ge 3k\Rightarrow n\ge k,\text{ absurd.}$$ Thus, we have $\mathrm{deg}Q(x)=0$, that is $Q(x)=c\ne 0$ and then $A(x)=B(x)$: $$\begin{gathered} 3(c+1)x^{2n}+3(c^2+2c+(-1{)}^n)x^n+c^3+3c^2+2c=0,\text{ for all }x\in \mathbb{R}\iff \\ c+1=0,c^2+2c+(-1{)}^n=0,c^3+3c^2+2c=0\iff c=-1,n=2m,m\in {\mathbb{N}}^{\ast }. \end{gathered}$$ Hence the polynomial: $P(x)=x^{2m}-1,x\in \mathbb{R},m\in {\mathbb{N}}^{\ast }$ is a solution.

II. $a=-1$. Then relation (2) becomes: $$\begin{gathered} (-x^n+Q(x){)}^3+3(-x^n+Q(x){)}^2=-x^{3n}+Q(x^3)-3Q(-x)+3(-1{)}^n{x}^n \\ \iff A(x)=B(x), \end{gathered}$$ where we have put $$\begin{gathered} A(x)=3x^{2n}Q(x)-3x^n(Q(x){)}^2+(Q(x){)}^3+3x^{2n}-6x^{n}Q(x)+3(Q(x){)}^2 \\ B(x)=Q(x^3)-3Q(-x)+3(-1{)}^n{x}^n. \end{gathered}$$ Working as in case (I), if $Q(x)=0$, then: $3x^{2n}=3\cdot (-1{)}^n{x}^n$, (impossible). If $\mathrm{deg}Q(x)=k>0$, then, since $0<k<n$, we have: $$2n+k=\mathrm{deg}A(x)=\mathrm{deg}B(x)=\max \{3k,n\}\ge 3k\Rightarrow n\ge k,\text{ absurd.}$$ Hence $\mathrm{deg}Q(x)=0$, that is $Q(x)=c\ne 0$ and then from equality $A(x)=B(x)$ we have: $$\begin{gathered} 3(c+1)x^{2n}-3(c^2+2c+(-1{)}^n)x^n+c^3+3c^2+2c=0,\text{ for all }x\in \mathbb{R} \\ c+1=0,c^2+2c+(-1{)}^n=0,c^3+3c^2+2c=0\iff c=-1,n=2m,m\in {\mathbb{N}}^{\ast }. \end{gathered}$$ Hence we have the solution: $P(x)=-x^{2m}-1,x\in \mathbb{R},m\in \mathbb{N}$.

Finally, all solutions of the problem are the following: $$P(x)=-1,P(x)=-2,P(x)=x^{2m}-1,P(x)=-x^{2m}-1,x\in \mathbb{R},m\in {\mathbb{N}}^{\ast }.$$