Problem 1918

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

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

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

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