Finding a minimal polynomial of an algebraic element using Galois theory There is a canonical (but difficult) way of determining the minimal polynomial of an algebraic element $\alpha$ in a field $F$, namely by considering the $F$-linear transformation defined by left multiplication by $\alpha$ and computing the minimal polynomial of the matrix of the linear transformation. 
I am interested in a Galois-theoretic way of doing this. My motivation is the following: consider the problem of finding the minimal polynomial of $\sqrt2 + \sqrt3$ over $\mathbb Q$. One can do this by "conjugate bashing" and checking that 
$$(x - (\sqrt2 + \sqrt3))(x - (\sqrt3-\sqrt2))(x - (\sqrt2 - \sqrt3))(x + (\sqrt2 + \sqrt3))$$
is irreducible over $\mathbb Q$. The conjugate bashing is the same thing as applying all the Galois elements. This naive approach fails with $\sqrt[3]{2}$, which has minimal polynomial $x^3 - 2$ and Galois group of order 6. Not all hope is lost, however, since this is a "characteristic polynomial" of the element and contains as a factor the minimal polynomial. 
Is there any way to fix this deficiency?
 A: There is no "deficiency". $X^3 - 2$ having a Galois group of order $6$ is not a problem. When you apply all the automorphisms from the Galois group things reappear. Ignore the reappearance and you are done.
But you don't really need the galois group of the minimal polynomial. What you are really doing is, you are looking at all the automorphisms of the algebraic closure and making a list of places your element is sent to. Call that set $S$, then your minimal polynomial is $\prod_{t \in S} (X - t)$.
In your first example, $\sqrt 2 \not \in \mathbb Q (\sqrt 3)$ and vice versa. So you can have an automorphism of $\overline {\mathbb Q}$ that sends $\sqrt 2 \to \pm \sqrt 2$ and $\sqrt 3 \to \pm \sqrt 3$. Therefore your set S consists of $\left \{\pm \sqrt 2 \pm \sqrt 3 \right \}$. 
Similarly deal with $\sqrt[3] 2$.
A: If $K/F$ is Galois and $\alpha \in K$, then the minimal polynomial of $\alpha$ has as its roots the elements in the orbit of $\alpha$ under the action of $G(K/F)$.  That is to say, if $\{a_k\}$ is the set of elements in the orbit of $\alpha,$ then $\displaystyle \min_{\alpha}(x) = \prod_{k} (x-a_k)$.
For more information, see the bottom of page 6 here.
A: Here’s a method I use that doesn’t involve so much conjugate bashing, but does require a bit of arithmetic, namely to check that $3$ is still square-free in the PID $\Bbb Z[\sqrt2\,]$.
Suppose, for variety, that you want the minimal polynomial of $\sqrt[3]3+\sqrt2$, over $\Bbb Q$. You do know the minimal polynomial of $\sqrt[3]3$ over $\Bbb Q(\sqrt2\,)$, namely $f(X)=X^3-3$, and thus you know the minimal polynomial of $\sqrt[3]3+\sqrt2$, over $\Bbb Q(\sqrt2\,)$, namely $g(X)=f(X-\sqrt2\,)$. Just multiply $g$ by its conjugate (replacing $\sqrt2$ by $-\sqrt2$) to get the desired polynomial over $\Bbb Q$.
