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As $latex P(x^3-1)$ is divisible by $latex P(x^2+x+1)$, there exists a polynomial $latex Q(x)$ such that $latex P(x^2+x+1)Q(x)=P(x^3-1)$. Let $latex z$ be a zero of $latex P$. If $latex x^2+x+1=z$ then $latex x=\frac{-1\pm\sqrt{1+4(z-1)}}{2}$, i.e. $latex x^3-1= (x-1)(x^2+x+1)=z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right)$. The given equation means that $latex z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right)$ are also zeroes of $latex P$.Claim One of the two numbers $latex \left|\frac{-3\pm\sqrt{4z-3}}{2}\right|$ is greater than 1. ProofSuppose that they are both $latex \le 1$. Then we have  $latex 2\ge |-3+\sqrt{4z-3}|$ $latex 2\ge |-3-\sqrt{4z-3}|$Adding these two inequalities we get $latex 4\ge |-3+\sqrt{4z-3}|+|-3-\sqrt{4z-3}|\ge |(-3+\sqrt{4z-3})+(-3-\sqrt{4z-3})|=6$ which is absurd. Hence at least one of them has to be greater than 1.  The claim means that, if $latex z$ is non-zero, then at least one of the two zeroes $latex z\left(\frac{-3\pm\sqrt{4z-3}}{2}\right)$  has absolute value greater than $latex |z|$.

Now let $latex z=z_0$, the zero of $latex P$ with the largest absolute value. The procedure mentioned above can be used to construct a zero with absolute value bigger than $latex |z_0|$, which is absurd unless $latex z_0=0$. As $latex z_0$ is largest among the zeroes in absolute value, this means that $latex P$ does not have any nontrivial zeroes. Hence $latex P(x)=ax^n$ for some $latex a\in\mathbb{R}$ and $latex n\in\mathbb{N}$.

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