Let $P(x)$ be an irreducible polynomial in $\mathbb{Q}[x]$ such that $\text{deg}(P)$ is odd. $Q(x),R(x)$ are polynomials with rational coefficients satisfying $$P(x)|Q(x)^2+Q(x)R(x)+R(x)^2.$$Prove that $$P(x)^2|Q(x)^2+Q(x)R(x)+R(x)^2.$$
I found this problem in aops.com but it’s in the old post which is now already locked and no one solved it yet in that post so I will post it here. I tried to use the fact that $P(x)$ has at least one real root and it is irreducible in $\mathbb{Q}$. Also $P(x)\mid Q(x)^3-R(x)^3$ look like a good thing to consider but I don’t know how to continue from here.
Btw. I only want some hints not a full solution.