I am studying lang undergraduate algebra, field theory where this example is given.
Now I understand how $E_1 \cap E_2=\mathbb{Q}$. As we have $ \mathbb{Q} \subset E_1 \cap E_2 \subset E_1$. And we have $[E=\mathbb{Q(\alpha)}: \mathbb{Q}]=3$ since the basis of vector space $E_1$ over $\mathbb{Q}$ is ${1,\alpha, \alpha^2}$. So now $[E_1:\mathbb{Q}]=[E_1 : E_1 \cap E_2][E_1 \cap E_2 :\mathbb{Q}] \implies 3= [E_1 : E_1 \cap E_2][E_1 \cap E_2 :\mathbb{Q}]$. So that's why $[E_1 : E_1 \cap E_2] \ \vert \ 3$, since $E_1 \neq E_2$ $[E_1 : E_1 \cap E_2] \neq 1$ so $[E_1 \cap E_2 :\mathbb{Q}] =1$ and $E_1 \cap E_2 =\mathbb{Q}$.
Now $\mathbb{Q} \subset E_1 \subset E_1E_2$ so $[E_1E_2 : \mathbb{Q}]=[E_1E_2:E_1][E_1:\mathbb{Q}]$. So $[E_1E_2:E_1] =\frac{[E_1E_2 : \mathbb{Q}]}{3}$.But at the end I don't understand how they come to conclusion that $[E_1E_2:E_1]=2$. How they have calculate $[E_1E_2=\mathbb{Q}(\alpha,\beta)=\mathbb{Q}(\alpha,\sqrt(-3) : \mathbb{Q}]=6$. Any help is appreciated thanks.
Since $\alpha$ has minimal polynomial $x^3-2$ and $\sqrt{-3}$ has the minimal polynomial $x^2+3$, the combined degree of the extension is $2\times 3=6$.
This calculation implicitly uses $\Bbb Q(\alpha,\sqrt{-3})\cong\Bbb Q(\alpha)(\sqrt{-3})$ and we compute the degrees one at a time:
$$6=\underset{2}{\underbrace{[\Bbb Q(\alpha,\sqrt{-3}):\Bbb Q(\alpha)]}}\cdot\underset{3}{\underbrace{[\Bbb Q(\alpha):\Bbb Q]}}$$