Calculate $[\mathbb{Q} (\sqrt[3]{3} + \sqrt {2}) : \mathbb{Q}]$ Good morning, 
I'd like to understand the following concept and possibly being able to solve different types of exercises with it.
The exercise is the following : 
Calculate $[\mathbb{Q} (\sqrt[3]{3} + \sqrt {2}) : \mathbb{Q}]$
The solution seems pretty standard except for the point that i would like to understand.
It says more or less that once called $\alpha = \sqrt[3]{3} , \beta = \sqrt {2}$ the set $\{ 1, \alpha, \alpha^{2} , \beta, \alpha\beta , \alpha^{2}\beta \}$ is a basis of $\mathbb{Q}(\alpha, \beta)$ over $\mathbb{Q}$, instead $A= \{ 1, \gamma \dots , \gamma^{4} \}$ where $\gamma = \alpha + \beta$ is linearly dependent over $\mathbb{Q}$.
So it writes the coordinates of $A$ respect to the basis of $\mathbb{Q}(\alpha, \beta)$ :
$1=(1,0,0,0,0,0)$
$\gamma = \alpha + \beta = (0,1,0,1,0,0)$
$\gamma^{2} = \alpha^{2} + 2\alpha\beta + 2 = (2,0,1,0,2,0)$, where $2 = \beta^{2}$
$\gamma^{3}=(3,6,0,2,0,3)$.
After that it says that we can stop at $\gamma^{3}$ because we can see passed in coordinates and so in $\mathbb{Q}^{6}$ that the four element are linearly independent and finishes saying that $\mathbb{Q}(\gamma) = \mathbb{Q}(\alpha, \beta)$ and writing down $\alpha$ and $\beta$ as linear combination of element of $\gamma$, which I din't understood at all, because starts to put coefficients in matrices.
I'd like to understand the method and when I could apply it,and the last part on the matrices.
Thank you all, any help would be appreciated.
P.S.
If you need the text of the solution,please comment it and let me know.
 A: We know that $[\mathbb{Q}(\alpha, \beta) :  \mathbb{Q}] = 6$ and $\mathbb{Q}(\gamma) \subseteq \mathbb{Q}(\alpha, \beta)$. Therefore
$$6 = [\mathbb{Q}(\alpha, \beta) : \mathbb{Q}] = [\mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\gamma)][\mathbb{Q}(\gamma) : \mathbb{Q}]$$
so $[\mathbb{Q}(\gamma) : \mathbb{Q}] \mid 6$.
Also you know that one basis for $\mathbb{Q}(\gamma)$ over $\mathbb{Q}$ is of the form $$\{1, \gamma, \gamma^2, \ldots, \gamma^{n-1}\}$$ where $n = [\mathbb{Q}(\gamma) : \mathbb{Q}]$.
You showed that $\{1, \gamma, \gamma^2, \gamma^3\}$ is linearly independent over $\mathbb{Q}$ so $[\mathbb{Q}(\gamma) : \mathbb{Q}] \ge 4$.
Therefore $[\mathbb{Q}(\gamma) : \mathbb{Q}] = 6$.
This now implies that $[\mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\gamma)] = 1$ so $\mathbb{Q}(\alpha, \beta) = \mathbb{Q}(\gamma)$. Hence, $\alpha$ and $\beta$ can be expressed in the basis $\{1, \gamma, \ldots, \gamma^5\}$.
A: I think the argument is just that
$$
\mathbb{Q}(\gamma) \subseteq \mathbb{Q}(\alpha, \beta),
\
[\mathbb{Q}(\gamma): \mathbb{Q}]\ge 4
\
[\mathbb{Q}(\alpha, \beta): \mathbb{Q}]=6,
\implies
[\mathbb{Q}(\gamma): \mathbb{Q}]=6
$$
