https://artofproblemsolving.com/community/c6h277424 see this that solution solves the problem....

Let $P(x^3-1)=P(x^2+x+1)Q(x)$. Suppose $P$ is non constant. If it has a nonzero root, let $a$ be the root with the largest absolute value. Consider the quadratic equation $x^2+x+1=a$. Pick a root $x_1$ of this equation such that $|x_1-1|>1$. To see why this is possible, note that if $|x_1-1|\le 1$, then the other root is $x_2=-1-x_1$, so that $|x_2-1|=|x_1+2|\ge 3-|x_1-1|\ge 2$. Then plugging $x_1$ into the first equation, $a(x_1-1)$ is a root of $P$ and has a bigger absolute value than $a,$ contradiction. So $0$ is the only root of $P$, so it's actually $cx^k$, which works. Constant polynomials all work, so all solutions are given by $P(x)=cx^k$ for real $c$ and nonnegative integer $k$. $\blacksquare$

Here is my solution. (Which I wrote in exam). Assume $P(x)$ is non constant. $P(x^2 + x + 1)Q(x) = P(x^3 - 1)$ Let $ P (\alpha) = 0 $. So taking $ x^2 + x + 1= \alpha $. We get $x^3 - 1 = (x-1)(x^2 + x+1)= (x-1)(\alpha)= (\alpha)(\frac{-3 \pm \sqrt{4\alpha - 3}}{2})$ is also a root. Let $\beta$ be a root with maximum absolute value. Then we have $|\beta| \ge |\beta||\frac{-3 \pm \sqrt{4\beta - 3}}{2}|$ ........(1) Assume $|\beta| >0$ Then we have by (1) we have. $2\ge |-3 + \sqrt{4\beta -3}|$.........(2) and $ 2\ge |-3 - \sqrt{4\beta -3}|$.........(3) Adding (2) and (3) we get, $4 \ge |-3 + \sqrt{4\beta-3}| + |-3 - \sqrt{4\beta -3}| \ge |-6|$ Where the last inequality follows from the triangle inequality. Which is a contradiction. Thus implies $|\beta|=0$ and thus all roots are 0. Thus $P(x)= cx^n$ for $n \in \mathbb {Z}$

AnArtist wrote: Here is my solution. Assume $P(x)$ is non constant. $P(x^2 + x + 1)Q(x) = P(x^3 - 1)$ Let $ P (\alpha) = 0 $. So taking $ x^2 + x + 1= \alpha $. We get $x^3 - 1 = (x-1)(x^2 + x+1)= (x-1)(\alpha)= (\alpha)(\frac{-3 \pm \sqrt{4\alpha - 3}}{2})$ is also a root. Let $\beta$ be a root with maximum absolute value. Then we have $|\beta| \ge |\beta||\frac{-3 \pm \sqrt{4\beta - 3}}{2}|$ ............(1) Assume $|\beta| >0$ Then we have by (1) we have. $2\ge |-3 + \sqrt{4\beta -3}|$.........(2) and $ 2\ge |-3 - \sqrt{4\beta -3}|$.........(3) Adding (2) and (3) we get, $4 \ge |-3 + \sqrt{4\beta-3}| + |-3 - \sqrt{4\beta -3}| \ge |-6|$ Where the last inequality follows from the triangle inequality. Which is a contradiction. Thus implies $|\beta|=0$ and thus all roots are 0. Thus $P(x)= cx^n$ for $n \in \mathbb {Z}$ $\alpha$ is a complex number...

@above. So what?

AnArtist wrote: @above. So what? You should avoid using $\sqrt{}$ signs then...norms instead of absolute values and so on...

bel.jad5 wrote: AnArtist wrote: @above. So what? You should avoid using $\sqrt{}$ signs then...norms instead of absolute values and so on... Not required. Roots of complex numbers are defined.

$\underline{\textcolor{green}{\text{Answer}}}$ I claim that $P(x) \equiv cx^n $. proof Without losing generality let assume \[P(x^3-1)=Q(x) P(x^2+x+1)\]and degree of polynomial $P$ is $n$. Now ,we can show that $P(0)=0$. this is because , \[ P(\omega ^3-1)=Q(\omega ) P(0)\]So we get , either \[P (0)=0\]otherwise \[Q(\omega)=Q(\omega ^2)=1\]. Now we should show that only possibility is $P(0)=0$ or equivalently $Q(\omega ),Q(\omega ^2)\neq 1$. suppose $\alpha$ is root of $P$ such that $\alpha \neq 0$. WLOG $x^2+x+1=\alpha$ have roots $x_1,x_2$ which forces that $P(x_1^3-1)=P(x_2^3-1)=0$. Note that \[x_1^3+x_2^3 =(x_1+x_2)\{(x_1+x_2)^2 -3x_2x_1\}\] Now from triangle inequality we have $|x_1^3-1|+|x_2^3-1|\ge |x_1^3+x_2^3-2|=3|\alpha|$ so we must get one of the roots are greater than $\frac {3 |\alpha |}{2} $ say $x_1^3-1$ is that root. \[ |x_1^3-1|>|\alpha| \] which means that if we interchange $\alpha$ by $x_1^3-1$then we get another root $\beta$ with $|\beta |>|x_1^3-1|$. So we get a sequence of infinite root. It forces$P(x)\equiv 0$. which is impossible since $P(x^2+x+1)\mid P(x^3-1)$. Which implies $P(x)=0$ has only root $0$ and hence \[ P(x)=cx^n ,c\in R, c\neq 0 \].

$P \equiv k$ where $k$ is a non-zero constant and $P(x)=cx^n$ where $c$ is a constant and $n$ is a positive integer are the only solutions. Both of them clearly work and we will now proceed to show that they are the only solutions. We will only look for non-constant solutions of $P$ as the constant ones are fairly trivial. We will mainly show that the only root of $P$ is 0. By the problem condition there exists $Q \in \mathbb{R}[x]$ such that $P(x^2+x+1)Q(x)=P(x^3-1)$. Call this ${\color{red}(\clubsuit)}$. Claim: If $\alpha$ is a root then so is $\alpha \left(\dfrac{-3 \pm \sqrt{4\alpha -3}}{2}\right)$. Proof: Put $x^2+x+1 =\alpha$ in ${\color{red}(\clubsuit)}$ which nicely constructs another root of $P$. This gives \[ x^3-1=\alpha(x-1) \]We aim to compute $x-1$ in terms of $\alpha$ as then we can have $x^3-1$ in terms of $\alpha$ which is another root of $P$ by simple substitution in ${\color{red}(\clubsuit)}$ . This can be done by putting $y=x-1$ in $x^2+x+1 =\alpha$. This proves the claim. $\square$ Let $\beta$ be a root of $P$ with maximum modulus. Then by the above claim we have \begin{align*} |\beta| \ge \Bigg|\dfrac{-3 - \sqrt{4\beta -3}}{2}\Bigg| \tag{1.1} \label{1.1}\\ |\beta| \ge \Bigg|\dfrac{-3 + \sqrt{4\beta -3}}{2}\Bigg| \tag{1.2} \label{1.2} \end{align*} Adding $(1.1)$ and $(1.2)$ we get \begin{align*} 2|\beta| &\ge \underbrace{\Bigg|\beta\left(\dfrac{-3 + \sqrt{4\beta -3}}{2}\right) \Bigg| + \Bigg|\beta\left(\dfrac{-3 - \sqrt{4\beta -3}}{2}\right) \Bigg| \ge \Bigg| \beta\left(\dfrac{-3 + \sqrt{4\beta -3}}{2}\right) + \beta\left(\dfrac{-3 - \sqrt{4\beta -3}}{2}\right) \Bigg|}_{\text{Triangle Inequality}} \\ &=3|\beta| \end{align*}Now, comparing the extreme sides of the inequality, we clearly get that $|\beta|=0$. As $\beta$ had the maximum modulus which is 0, this implies that all roots of $P$ must be 0 and therefore we are done. $\blacksquare$

Let $t$ be a root of $P(x)=0$. Then we construct the sequence $\{t_n\}_{n \ge 0}$ going by $t_{n+1}=t(\frac{\pm \sqrt{4t-3}-3}{2})$. and $t_0=t$. So we get infinitely many roots or we get that the sequence is periodic. if the sequence turns out to be periodic, Let $|T|=\max\{|t_n| : n\ge 0\}$. Then $|T| \ge |T||\frac{\sqrt{4t-3}-3}{2}|$. $|T| \ge |T||\frac{-\sqrt{4t-3}-3}{2}|$ Cancelling $|T|$ from both sides assuming $|T| \neq 0$ gives that $2 \ge |\sqrt{4t-3}-3|$ $2 \ge |-\sqrt{4t-3}-3|$ Adding them we have $4 \ge |\sqrt{4t-3}-3|+|-\sqrt{4t-3}-3| \ge |-6|$ $\implies 4 \ge 6$ which is a "flying cow"( ) statement. So $|T|=0$ is maximal. Thus $P$ has no root except $0$. Taking multiplicity of $0$ to be $n$ and $c$ a real number, we get $P(x)=cx^n$.

ftheftics wrote: $\underline{\textcolor{green}{\text{Answer}}}$ I claim that $P(x) \equiv cx^n $. proof Without losing generality let assume \[P(x^3-1)=Q(x) P(x^2+x+1)\]and degree of polynomial $P$ is $n$. Now ,we can show that $P(0)=0$. this is because , \[ P(\omega ^3-1)=Q(\omega ) P(0)\]So we get , either \[P (0)=0\]otherwise \[Q(\omega)=Q(\omega ^2)=1\]. Now we should show that only possibility is $P(0)=0$ or equivalently $Q(\omega ),Q(\omega ^2)\neq 1$. suppose $\alpha$ is root of $P$ such that $\alpha \neq 0$. WLOG $x^2+x+1=\alpha$ have roots $x_1,x_2$ which forces that $P(x_1^3-1)=P(x_2^3-1)=0$. Note that \[x_1^3+x_2^3 =(x_1+x_2)\{(x_1+x_2)^2 -3x_2x_1\}\] Now from triangle inequality we have $|x_1^3-1|+|x_2^3-1|\ge |x_1^3+x_2^3-2|=3|\alpha|$ so we must get one of the roots are greater than $\frac {3 |\alpha |}{2} $ say $x_1^3-1$ is that root. \[ |x_1^3-1|>|\alpha| \] which means that if we interchange $\alpha$ by $x_1^3-1$then we get another root $\beta$ with $|\beta |>|x_1^3-1|$. So we get a sequence of infinite root. It forces$P(x)\equiv 0$. which is impossible since $P(x^2+x+1)\mid P(x^3-1)$. Which implies $P(x)=0$ has only root $0$ and hence \[ P(x)=cx^n ,c\in R, c\neq 0 \]. $\omega $ is not real.I think wrong solution.