= Solution
Squaring and eliminating the inner radical gives
$$
(\alpha^2-3)^2=3,\qquad
(\gamma^2-3)^2=6.
$$
Thus $\alpha,\beta$ are roots of
$$
X^4-6X^2+6,
$$
and $\gamma,\delta$ are roots of
$$
X^4-6X^2+3.
$$
Both polynomials are irreducible by the <Eisenstein criterion> at $3$, so these are the four <minimal polynomial>s over $\mathbb Q$.
Over $K=\mathbb Q(\sqrt3)$, the relevant radicands are
$$
a=3+\sqrt3,\qquad b=3-\sqrt3,\qquad ab=6.
$$
The norms of $a$ and $b$ to $\mathbb Q$ equal $6$, so neither can be a square in $K$. Also $(u+v\sqrt3)^2=6$ with $u,v\in\mathbb Q$ would force $uv=0$, followed by either $u^2=6$ or $v^2=2$, both impossible. Thus $6$ is not a square either.
Similarly, over $K'=\mathbb Q(\sqrt6)$ the radicands $3+\sqrt6$ and $3-\sqrt6$ have nonsquare norm $3$, while $(u+v\sqrt6)^2=3$ would force either $u^2=3$ or $v^2=1/2$. Hence $3$ is not a square in $K'$. Both extensions therefore satisfy all the hypotheses of part (b).
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