= Solution
{c}
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> gives
$$
\boxed{[\mathbb Q(\zeta_{18}):\mathbb Q]=6.}
$$
In fact this is the same <field> as in part (i): $\zeta_9=\zeta_{18}^2$ and $\zeta_{18}=-\zeta_9^5$.
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