 Solution

Squaring and eliminating the inner radical givesThus $\alpha ,\beta$ are roots ofand $\gamma ,\delta$ are roots ofBoth polynomials are irreducible by the Eisenstein criterion at $3$, so these are the four minimal polynomials over $\mathbb{Q}$.

Over $K=\mathbb{Q}(\sqrt{3})$, the relevant radicands areThe norms of $a$ and $b$ to $\mathbb{Q}$ equal $6$, so neither can be a square in $K$. Also $(u+v\sqrt{3}{)}^2=6$ 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 $K^{\prime }=\mathbb{Q}(\sqrt{6})$ the radicands $3+\sqrt{6}$ and $3-\sqrt{6}$ have nonsquare norm $3$, while $(u+v\sqrt{6}{)}^2=3$ would force either $u^2=3$ or $v^2=1/2$. Hence $3$ is not a square in $K^{\prime }$. Both extensions therefore satisfy all the hypotheses of part (b).