I have seen couple questions related to this one, but after reading the answers I am still confused:
Why is the extension dimension of $x^3 - 2$ equal to $6$? In other words, why are the basis $1,\sqrt[3]2,\sqrt[3]4,ω,ω\sqrt[3]2,ω\sqrt[3]4$? What happens to $ω^2,ω^2\sqrt[3]2$ and $ω^2\sqrt[3]4$? Why are they not basis elements as well? If the extension dimension is equal to the order of $G$, then is there a one-to-one correspondence between the basis and the elements of $G$?
Hint If $\omega$ is a nontrivial cube root of unity, then $\omega^3 - 1 = 0$ but $\omega - 1 \neq 0$. Thus, $$0 = \frac{\omega^3 - 1}{\omega - 1} = \omega^2 + \omega + 1,$$ and rearranging gives $$\omega^2 = -\omega - 1.$$