The polynomials factor as $(x+2)(x^2+x+1)$ and $(x-2)(x^2-x+1)$, respectively. So $1$ is a factor of $P$, as is $-2$, as is $\omega$ and $\omega^2$. Let $P(x)=(x-1)(x+2)(x^2+x+1)Q(x)$, and get $(x+2)(x^2+x+1)(x-2)(x+1)(x^2-x+1)Q(x-1)=$ $(x-2)(x^2-x+1)(x-1)(x+2)(x^2+x+1)Q(x)$ or $(x+1)Q(x-1)=(x-1)Q(x)$. Thus $-1$ and $0$ are roots of $Q(x)$; let $Q(x)=x(x+1)R(x)$ to get $R(x)=R(x-1)$ or $R(x)\equiv c$. Thus $P(x)=c(x-1)x(x+1)(x+2)(x^2+x+1)$ for any real $c$. We can easily check that this works.

\[ (x^{3}+3x^{2}+3x+2)P(x-1)=(x^{3}-3x^{2}+3x-2)P(x) \] \[(x+2)(x^2+x+1)P(x-1) = (x-2)(x^2-x-1)P(x) \implies P(1) = P(-2) = 0 \implies \] \[(x^3+1)(x+2)(x-2)G(x-1) = (x^3+1)(x+2)(x-2)G(x)\] \[(x^3+1)G(x-1) = (x^3-1)G(x) \implies G(0)= G(-1) = 0 \implies \] \[(x^3+1)(x-1)x M(x-1) = (x^3-1)x(x+1)M(x)\] \[(x^2+x+1)M(x-1) = (x^2-x+1)M(x)\] \[ M(x-1) = \frac{(x^2+x+1)M(x)}{x^2+x+1} \implies M(x) = Q(x)(x^2+x+1)\] \[(x^2+x+1)(x^2-x+1)Q(x-1) = (x^2-x+1)(x^2+x+1)Q(x)\] \[Q(x-1)= Q(x) \implies Q(x) = c\] $\implies P(x) = c(x^2+x+1)x(x+1)(x-1)(x+2)$