The kernel of this evaluation map. Consider $$f: \mathbb{Z}[x] \longrightarrow \mathbb{Z}[2^{1/3}]$$ which is a evaluation map for $x = 2^{1/3}$.
Question: What is the kernel of $f$?
My work so far: I take the ideal $(x^3-2)$ as my answer, and I want to show that $\ker(f) = (x^3-2)$. One side is trivial, i.e., $(x^3-2) \subset \ker(f)$. 
Any idea for the another inclusion?
Note that: $a_0 + 2a_3 + 4a_6 + 8a_9 + \dots = 0$, $a_1 + 2a_4 + 4a_7 + 8a_10 + \dots = 0$ and $a_2 + 2a_5 + 4a_8 + 8a_{11} + \dots = 0$ for any polynomial $p(x) = a_0 +a_1x + a_2x^2 + \dots \in \ker(f)$. This can help but the proof is not very clean (or not a valid proof at all), i.e., I need to deal with different cases for $n=3k$, $n=3k-1$ or $n=3k-2$.
Well, thanks for the help!

Hint:
The kernel has to be a non-maximal prime ideal in  $\mathbf Z[x]$, and  $x^3-2$ is irreducible in  $\mathbf Z[x]$, hence, as it is a U.F.D., the ideal $x^3-2$ is prime, and  not maximal. Furthermore,  $\mathbf Z[x]$ has Krull dimension $2$, so non-zero, non-maximal prime ideals all have height $1$.
