Why $[\mathbb Q( \omega, 2^{1/3}) : \mathbb Q]$ =6? 
Why  $[\mathbb Q( \omega, 2^{1/3}) : \mathbb Q]$ =6 ?, 

We know that the dimension of $\mathbb Q(\omega,2^{1/3})$ over $\mathbb Q$ will be equal to the degree of the minimal polynomial which has roots $\omega, 2^{1/3}$ .
Now $(x^2+x+1)(x^3-2)$ which has roots $\omega, 2^{1/3}$  so I thought the dimension should be 5. 
But in the computation $[\mathbb Q(\omega,2^{1/3}):\mathbb Q]=[\mathbb Q(\omega,2^{1/3}):\mathbb Q(\omega)][\mathbb Q(\omega): \mathbb Q] =6$.
Where am I wrong ? Please someone check.. Thank you. 
 A: The "dimension is equal to the degree of the minimal polynomial" thing only works when you have a single element that generates the whole extension. Take, for instance, $\alpha = \omega + 2^{1/3}$. It has minimal polynomial
$$
x^6 - 3 x^5 + 6 x^4 - 11 x^3 + 12 x^2 + 3 x + 1
$$
over $\Bbb Q$ and therefore generates an extension of degree $6$. Clearly $\Bbb Q(\alpha)\subseteq \Bbb Q(\omega, 2^{1/3})$, so the degree of $Q(\omega, 2^{1/3})$ cannot be $5$.
Using this, and the multiplicative formula for the extension degrees which you've already used in your question, we can see that if you have several distinct elements that generate a field extension, we multiply the degrees of their respective minimal polynomials instead of adding them.
Or... almost. We multiply the degree of the minimal polynomial of $\omega$ over $\Bbb Q$ with the degree of the minimal polynomial of $2^{1/3}$ over $\Bbb Q(\omega)$ (or vice versa). It might not make a difference in this case, but it is important with other extensions like $\Bbb Q(\sqrt2, \sqrt[4]2+\sqrt2)$, and many other, probably less trivial examples.
A: "We know that the dimension of $\mathbb Q(\omega,2^{1/3})$ over $\mathbb Q$ will be equal to the degree of the minimal polynomial which has roots $\omega, 2^{1/3}$ ."
This is false, and your question is a counterexample to this statement. The degree is equal to $6$, as you correctly derived from the multiplicative formula for the degrees.
