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 an answer, but I'm posting it to show why you want to follow the link and use the suggestions there, compared to how much pain attacking this the brute force method would give (as well as not really going anywhere.....)
Let's see...take an arbitrary polynomial, $$f(x)=\sum _{i=0}^na_ix^i $$ and see where we get plugging in terms
$$f(x^3-2)=\sum _{i=0}^na_i(x^3-2)^i=\sum _{i=0}^na_i\big(\sum _{j=0}^i\big (\overset i j\big)x^j(-2)^{i-j}\big)$$ whereas $$f(x)^3-2=f(x)=\big(\sum _{i=0}^na_ix^i\big)^3-2=\sum_{\sum_{k=o}^nm_k=3}\frac {3!}{\prod_{j=0}^nm_j!}\prod_{l=0}^n(a_lx)^{m_k}-2$$, the latter taken from the multinomial theorem (http://www.trans4mind.com/personal_development/mathematics/series/multiNomialTheorem.htm)
Since the power is 3, I'll rewrite that messy equation which is really just saying the sum of the powers is 3 into parts where we have a single power of 3, a power of 2 and a power of 1, and 3 powers of 1
Doing so, we get $$f(x)^3-2=\sum _{k=0}^n(a_kx^k)^3+\sum _{k=0}^n\prod_{i\ne k \text {from} 0}^n3(a_kx^k)^2a_ix^i+\sum_{i,j,k \text {all different,}=0}^n6a_ia_ja_kx^{i+j+k}-2$$
So, looking at the constant term, on the left hand side, equating the left hand side to the right hand side, we get $$\sum_{i=0}^n a_i(-2)^i=a_0^3-2$$, Attacking the lead term, again LHS to RHS, we get $$a_n^3=a_n^3$$, which gets us nowhere.
About 3 sentences ago, I noticed the link including the solutions, so I stopped here...but I thought posterity would be served by posting this to show brute force is not the best approach :)
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 $P(x)=Q(x)$.
