= Solution
Let $a=3^{1/4}>0$. The four roots are $a,ia,-a,-ia$, so \b[$K=\mathbb Q(a,i)$]. Eisenstein's criterion at three gives $[\mathbb Q(a):\mathbb Q]=4$. This field is real, so adjoining $i$ doubles the degree and $[K:\mathbb Q]=8$.
Define $r(a)=ia$, $r(i)=i$, and complex conjugation $s(a)=a$, $s(i)=-i$. The first automorphism exists because over $\mathbb Q(i)$ the minimal <polynomial> of $a$ has degree four and all four roots are in $K$; the second is immediate. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^js$ are distinct and exhaust the eight automorphisms. Thus
$$
\boxed{\operatorname{Gal}(K/\mathbb Q)\cong D_8,}
$$
the dihedral group of order eight, acting on the four roots as the symmetries of a square. On every element of $K$, the action is determined by the displayed images of $a,i$, or by extending them linearly to the basis $a^ji^e$ with $0\leq j<4$, $0\leq e<2$.
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