Minimal polynomial of $\sqrt[3]{2} + \sqrt{3}$ 
Suppose I want to find the minimal polynomial of the number $\sqrt[3]{2} + \sqrt{3}$.

Now that means I want to find a  unique polynomial that is irreducible over $\Bbb Q$ such that $f(x)=0$. Now I know that because   $\sqrt[3]{2} + \sqrt{3}$ belongs to $\Bbb Q( \sqrt[3]{2} , \sqrt{3})$  it might be degree $2$, $3$ or $6$ and does not belong to $\Bbb Q( \sqrt[3]{2})$ so it cannot be of degree $3$ or  $\Bbb Q( \sqrt{3})$ so it cannot be of degree $2$.
So it is of degree $6$. I think my sayings are a bit intuitive and not formal and lack rigorous.
Couldn't it belong to another extension of degree $2$? or $3$? I can't answer that. Why checking only those $2$ is enough? Or is it wrong at all to say that?

Let $a=\sqrt[3]{2} + \sqrt{3}$. Notice that 
$$(a-\sqrt{3})^3=2=a^3-3\sqrt 3 a^2+9a-3\sqrt 3 
= a^3+9a-\sqrt 3 (3a^2+3)  \tag 1$$
therefore
$$\sqrt 3 = \frac{a^3+9a-2}{3a^2+3}     \tag 2$$
In particular, $\Bbb Q(a)$ contains $\Bbb Q(\sqrt 3)$ and also contains $\Bbb Q(a-\sqrt 3) = \Bbb Q(\sqrt[3]{2})$.
Therefore your intuition is correct: the degree of $\Bbb Q(a)$ is a multiple of $3$ and a multiple of $2$ (over $\Bbb Q$).
The degree of the minimal polynomial of $a$ over $\Bbb Q$ is then at 
least $6$.
From $$(a^3+9a-2)^2 = [\sqrt 3 (3a^2+3)]^2     \tag 3$$
you get a monic polynomial $P \in \Bbb Q[X]$ of degree $6$, such that $P(a)=0$. Thus $P$ is the minimal polynomial of $a$ over $\Bbb Q$.
Here is the minimal polynomial $P(X)$ of $a$ over $\Bbb Q$ :

 $P(x) = x^6-9 x^4-4 x^3+27 x^2-36 x-23$.

