Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / A / 1 / ii

Replacing $x$ by $-x$ in the irreducible polynomial from part (i) gives $x^6-x^3+1$. This substitution is an invertible automorphism of $\mathbb{Q}[x]$, so it preserves irreducibility. The roots are the primitive eighteenth roots, and the splitting field is $\mathbb{Q}({\zeta }_{18})$. They are all powers of one such root, so its irreducible degree-six polynomial givesIn fact this is the same field as in part (i): ${\zeta }_9={\zeta }_{18}^2$ and ${\zeta }_{18}=-{\zeta }_9^5$.