Let's consider the polynomial $f=x^6+3 \in \Bbb Q[x]$. I have to prove that for some root $\beta $ of $f$. The extension $ \Bbb Q (\beta) $ is galois.
In other words if $ {\root 6 \of { - 3} } $ denotes some root of $f$. Then the extension $ {\Bbb Q}\left( {\root 6 \of { - 3} } \right)/{\Bbb Q} $ Is Galois
What I tried
Let's denote $ w = e^{\frac{i\pi}{3}}$ i.e the primitive 6-root of unity $w^6=1$. Let's denote $a_0 = \root 6 \of 3 e^{\frac{{i\pi }} {6}} $. The roots of $f$ are $ a_k = \root 6 \of { - 3} w^k \,\,\,k = 0..5 $ Given $a_0$ I want to generate all the roots, If I generate $a_1$ I'm done, but I don't know how. Here are some of my computations: $ \eqalign{ & a_0 = \frac{{\root 6 \of 3 }} {2}\left( {\sqrt 3 + i} \right) \cr & a_0 ^2 = \frac{{\root 3 \of 3 }} {2}\left( {1 + i\sqrt 3 } \right) \cr & a_0 ^3 = 8i\, \cr} $ I want to generate $ a_1 = a_0 w = \frac{{\root 6 \of 3 }} {2}\left( {\sqrt 3 + i} \right)\frac{1} {2}e^{\frac{{i\pi }} {3}} = i\root 6 \of 3 $ . How can I do it?