Is there a relation between the polynomial that generates a splitting field and its minimal polynomial? It's not too hard to show that the splitting field of $f(x) = x^3 - 2 \in \mathbb{Q}[x]$ over $\mathbb{Q}$ is $\mathbb{Q}(2^{1/3}, \omega)$ where $\omega$ is a nontrivial third root of unity (a little working shows that it can also be written as $\mathbb{Q}(2^{1/3}, \sqrt{-3})$). However, as one expects of a degree 6 extension, the minimal polynomial is also of degree 6: $x^6 + 9x^4 - 4x^3 + 27x^2 + 36x + 31$. Is there any relation between this higher degree minimal polynomial and the third degree polynomial we started with, considering they both generate the same splitting field? This is meant to be a general question, the above is just an example.
EDIT: A clarification and generalisation of the question. Apologies for the earlier lack of clarity, I see that I made too many assumptions and such.
Given a field $K$ and a polynomial $f(x)\in K[x]$, supposing we can find a field extension $K(\alpha_1, ..., \alpha_n) \cong K[x]/(f)$, is there any relation between $f$ and the minimal polynomial of some linear combination of the $\alpha_i$ (that involves all $\alpha_i$.
The reason I ask this is that I would expect there to be something relating them as both the original polynomial and the resulting minimal polynomial will result in the aforementioned field extension.
 A: If $K$ is a field of characteristic zero, if $\alpha_1,\dots,\alpha_n$ are algebraic over $K$, if there is no proper subset $S$ of $\{\,\alpha_1,\dots,\alpha_n\,\}$ such that $K(S)=K(\alpha_1,\dots,\alpha_n)$, then there is some linear combination $\alpha=\sum c_i\alpha_i$ such that $K(\alpha_1,\dots,\alpha_n)=K(\alpha)=K[x]/(f)$, where $f$ is the minimal polynomial for $\alpha$. 
In the other direction, if you start with $f$ irreducible over $K$, then there will be lots of ways to find $\alpha_1,\dots,\alpha_n$ such that $K(\alpha_1,\dots,\alpha_n)=K[x]/(f)$, and there's no reason to think there's a linear combination $\alpha=\sum c_i\alpha_i$ such that the minimal polynomial of $\alpha$ is related to $f$ (although there's a certain amount of wiggle room here, since it's not entirely clear what you mean by "any relation"). 
Consider this example. Let $K$ be the rationals, let $f$ be the minimal polynomial for $\sqrt2+\sqrt6$, then $K(\sqrt2,\sqrt3)=K[x]/(f)$, but there is no rational linear combination of $\sqrt2$ and $\sqrt3$ with minimal polynomial $f$. 
