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Note that $x^6+x^3+1$ is a factor of $x^9-1$, so I computed its roots are $$w,w^2,w^4,w^5,w^7,w^8$$where $w=e^{\frac{2\pi}{9}i}$. So the splitting field is just $Q(w)$. But I'm having trouble to find out what it is and its degree.Any hints?Thx

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up vote 3 down vote accepted

Can you prove $x^6+x^3+1$ irreducible over the rationals? If so, do you see how that answers your question?

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The multiplicative group generated by $\omega$ is cyclic of order $9$, and they are all roots of $x^9-1$. So the Galois group acts like a group of automorphisms of $\mathbf{Z}_9$; the action is faithful, because it is completely determined by what it does to $\omega$, and if a Galois automorphism fixes $\omega$ then it fixes $\mathbb{Q}(\omega)$.

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