If $P(x)^3=P^*(x^3)$ for some $P^*$, then $P(x)=a_nx^n+a_{n-3}x^{n-3}+\dots$, because if we look at the coefficient of $x^{3n-k}$ in $P^*$, it is $0$ if $k\equiv 1,2\pmod 3$, and it is equal to $3a_n^2a_{n-k}$, if $a_{n-1},a_{n-2},a_{n-4},a_{n-5},\dots$ are all zero before $a_{n-k}$ So, if $n$ is the degree of $P$ we can write \[P(x) = x^r\cdot Q(x^3),\text{\, \, where\,}r=n\mod 3 \]Now, we can solve easily.

Satyaprakash2009rta wrote: If $P(x)^3=P^*(x^3)$ for some $P^*$, then $P(x)=a_nx^n+a_{n-3}x^{n-3}+\dots$, because if we look at the coefficient of $x^{3n-k}$ in $P^*$, it is $0$ if $k\equiv 1,2\pmod 3$, and it is equal to $3a_n^2a_{n-k}$, if $a_{n-1},a_{n-2},a_{n-4},a_{n-5},\dots$ are all zero before $a_{n-k}$ So, if $n$ is the degree of $P$ we can write \[P(x) = x^r\cdot Q(x^3),\text{\, \, where\,}r=n\mod 3 \]Now, we can solve easily. Even though this was the main part, it is still not trivial after this.

YESMAths wrote: Even though this was the main part, it is still not trivial after this. Here is a similar problem. check out problem 6 here: http://www.imomath.com/index.php?options=346&lmm=0

Any solution

Satyaprakash2009rta wrote: YESMAths wrote: Even though this was the main part, it is still not trivial after this. Here is a similar problem. check out problem 6 here: http://www.imomath.com/index.php?options=346&lmm=0 The way it is done there, the polynomials happen to be quadratics and so the other non-trivial case is manipulated to find an infinite sequence where the polynomial gives the same value. But in this case, it becomes a bit nasty to deal with $P(x)=x^2Q(x^3-2)$, and that's why the problem ain't so trivial after the main lemma. Moreover, I suppose that the $r$ in your first post can be reduced mod $3$ to get the possibilities $r=0, 1, 2$ ($r=0$ is easy to deal with and $r=1$ is also manipulated to create an infinite sequence of integers at which $Q$ assumes the same constant value).

The main thing is to see that $P$ is a polynomial in $x^3$ or $x$ times $P$ or $x^2$ times $P$ is. Then it's trivial.

@anantmudgal can you post your solution please? I think its not easy by coeff comparing.

Anant is right.See that LHS is a pol in $x^3$ so $P(x)$ is either $Q(x^3),xQ(x^3),x^2Q(x^3)$,by coeff comparing.The latter two imply $P(0)=0$ where you easily get $P(x)=x$ as only solution.Else,you apply a method similar to infinite descent to show that all other polynomials are $P(P(....P(x)...))$ with $P(x)=x^3-2$

@above You are right. I will show that if $ f( 0) =0$, then $ f( x) =x$ (even if it is easy) Let $ ( a_{n})$ be the sequence defined by $ a_{0} =0$, $ a_{n+1} =\sqrt[3]{2+a_{n}}$ recursively. We will show that $ a_{n+1} >a_{n}$ for each $ n\in \mathbb{N}_{\geq 0}$ by induction on $ n$. For $ n=0$, we have $ a_{1} =\sqrt[3]{2} >0=a_{0}$. Suppose that $ a_{n+1} >a_{n}$ for some $ n$. $ a_{n+1} >a_{n} \ \Rightarrow \ \sqrt[3]{2+a_{n+1}} >\sqrt[3]{2+a_{n}} \ \Rightarrow \ a_{n+2} >a_{n+1}$. So we can say that $ a_{n+1} >a_{n}$ holds for each $ n\in \mathbb{N}_{\geq 0}$ Let $ g( x) =f( x) -x$. Suppose that $ g( a_{n}) =0$ for some $ n$. We have $ a_{n} =f( a_{n}) =f\left( a^{3}_{n+1} -2\right) =f( a_{n+1})^{3} -2$. Since $ f( x) \in \mathbb{R}[ x]$ and $ a_{n} ,a_{n+1} \in \mathbb{R}$, we get $ f( a_{n+1}) =a_{n+1}$ from the above relation. It follows that $ g( a_{n+1}) =0$ , of course. Thus, $ g( x)$ is indeed zero polynomial. Therefore, if $ f( 0) =0$, then $ f( x) =x$.