Past exam of the options I course of the University of Oxford / 2009 / A / A1 / i / Solution

The polynomial identityshows that the roots are the ninth roots of unity which are not third roots. These are precisely the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus the splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility statement makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving