= Solution
A primitive sixth root of unity satisfies
$$
\boxed{\Phi_6(x)=x^2-x+1,}
$$
which is irreducible over $\mathbb Q$ because its discriminant is $-3$. For the other polynomial choose $\alpha=i\,3^{1/6}$, so $\alpha^6=-3$ and $\alpha^3=-i\sqrt3$. Then $\zeta_6=(1-\alpha^3)/2\in\mathbb Q(\alpha)$, so this field already contains all six roots. The <Eisenstein criterion> at $3$ proves $x^6+3$ irreducible, giving a <splitting field> of degree six.
Put $r(\alpha)=\zeta_6^2\alpha$ and $s(\alpha)=\zeta_6\alpha$. The first automorphism fixes $\zeta_6$ and has order three. The second sends $\alpha^3$ to its negative, so sends $\zeta_6$ to $\zeta_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the <Galois group>. Therefore
$$
\boxed{\operatorname{Gal}(x^6+3/\mathbb Q)\cong S_3.}
$$
The presence of sixth roots of unity inside the <splitting field> is why this particular binomial has degree six rather than a generic twelve-degree <splitting field>.
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