Degree of field extension using minimal polynomials

Question

My notes say that $[\mathbb{Q}(\omega,2^{1/3}): \mathbb{Q}]$ equals 6.

How is this so? We have

$[\mathbb{Q}(\omega,2^{1/3}): \mathbb{Q}]=[\mathbb{Q}(\omega,2^{1/3}):\mathbb{Q}(2^{1/3})][\mathbb{Q}(2^{1/3}):\mathbb{Q}]= \deg(m_{\mathbb{Q}(2^{1/3})}(\omega)) \cdot \deg(m_{\mathbb{Q}}(2^{1/3}))=\deg(x^3 -1)\cdot \deg(x^3-2) = 9 $

?

Answer 1

Note that $x^3-1=(x-1)(x^2+x+1)$.

$\deg(m_{\mathbb{Q}(2^{1/3})}(\omega))=\color{red}2 ,$ since $\omega^2+\omega+1=0.$