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!

As you said in the comments, $2^\frac13$ is a root of $p\in \mathbb Q[x]$ if and only if the minimal polynomial of $2^\frac13$ divides $p$ in $\mathbb Q[x]$.
Why is $x^3-2$ the minimal polynomial? Remember, the minimal polynomial is the monic rational polynomial of smallest degree that has $2^\frac13$ as a root. So, is $2^\frac13$ the root of any first- or second-degree rational polynomial? No, so $x^3-2$ is the minimal polynomial. 
Alternatively, if we can show that $x^3-2$ is irreducible, then we also know that it is the minimal polynomial. Since it is of degree $\le 3$, it suffices to show that it has no rational roots, which is true because the only real root is $2^\frac13$. If you know Eisenstein's criterion, that also works here.
Now we can finish. Let $p\in\mathbb Z[x]$. Then
$$
p \in \ker f
\iff p(2^\frac13)=0
\iff p = q\cdot(x^3-2),\ q\in\mathbb Q[x]
$$
Remember that $p$ has integer coefficients. This forces $q$ to be in $\mathbb Z[x]$ as well. If not, then let $cx^k$ be the highest-degree term in $q$ where $c\notin\mathbb Z$. Then the $(k+3)$-degree term in $p$ will not be an integer; contradiction.
So we are done:
$$
p \in \ker f \iff p = q\cdot(x^3-2),\ q\in\mathbb Z[x] \iff p\in (x^3-2)
$$
