Real polynomials with $P(x^3-2)=P(x)^3-2$ Find all real polynomials $P(x)$ such that $P(x^3-2)=P(x)^3-2$.
Clearly $P(x)=x$ works. If $P(x)=ax+b$ is linear, then $P(x^3-2)=ax^3-2a+b$ and $P(x)^3-2=a^3x^3+3a^2bx^2+3ab^2x+b^3-2$, so $ab=0$ and $a^3=a$ and $-2a+b=b^3-2$. 
If $b=0$, then the last equation implies that $a=1$. 
If $a=0$, then the last equation implies that $b^3-b-2=0$, which has a real root and yields a constant solution.

This is not a complete answer as well but I hope the following result will help. This result guarantees there exist exactly one solution for polynomials of degree which is a power of $3$. If one can prove there exist no solution for polynomials with degree not a power of $3$ then the whole proof can be completed.
Lemma: If $P(x)$ and $Q(x)$ has same degree and $P(x^3-2)=P(x)^3-2$ and $Q(x^3-2)=Q(x)^3-2$ then $P(x)=Q(x)$.
Let $P(x)=\Sigma_0^n a_ix^i$ and let $Q(x)=\Sigma_0^nb_ix^i$.
Then $P(x^3-2)-Q(x^3-2)=P(x)^3-Q(x)^3$.
Base case: consider the coefficient of $x^{3n}$ on both side, left side has $a_n-b_n$ and right side has ${a_n}^3-{b_n}^3$ so $a_n-b_n=(a_n-b_n)({a_n}^2+a_nb_n+{b_n}^2)\implies a_n=b_n$.
Now suppose $a_i=b_i$ for all $i=n,n-1,n-2,...,k$. We look at the case $a_{k-1}$ and $b_{k-1}$.
First all the power of $x$ which is not a multiple of $3$ on the left side is obviously $0$. Now we look at the coefficient of $x^{3k-3}$ on the left side. By binomial theorem expansion it equals ${k-1\choose 0}(a_{k-1}-b_{k-1})+{k\choose 1}(a_k-b_k)+{k+1\choose2}(a_{k+1}-b_{k+1})+...+{n\choose n-k-1}(a_n-b_n)$.
By our induction hypothesis all terms except the first term is $0$ so it is equal to ${k-1\choose 0}(a_{k-1}-b_{k-1})$.
Suppose $a_{k-1}-b_{k-1}\neq 0$ then the left side has degree $3k-3$
Now right side is $P(x)^3-Q(x)^3=(P(x)-Q(x))(P(x)^2+P(x)Q(x)+Q(x)^2)$ has degree $(k-1)+2n=2n+k-1\geq 3k-1>3k-3$ contradiction.
Hence $a_{k-1}=b_{k-1}$ and by induction, we have all $a_i=b_i$ and hence $P(x)=Q(x)$.
