 Solution

For an irreducible cubic over $\mathbb{Q}$, the Galois group is $A_3$ when its discriminant is a rational square and $S_3$ otherwise. The rational-root test shows that the first two cubics are irreducible.

For $X^3-21X-22$,is not a square, so the Galois group is $S_3$.

For $X^3-21X-28$,so the Galois group is $A_3\cong C_3$.

Finally,whose remaining roots are $1\pm 3\sqrt{2}$. Its splitting field is $\mathbb{Q}(\sqrt{2})$ and its Galois group is $C_2$.