Galois group of ($x^2$ + $2x$ − $1$)($x^3$ + $3x^2$ + $3x$ − 6) over $\Bbb Q$

Question

What is the Galois group of $(x^2 + 2x − 1)(x^3 + 3x^2 + 3x − 6)$ over $\mathbb Q$?

I noticed that this polynomial can also be written as $((x+1)^2-2)((x+1)^3-7)$, can this be of any help?

Can I treat the polynomial as $((x')^2-2)((x')^3-7)$ for $x'=x+1$ and would that change the answer?

Answer 1

Both things you've noticed is of help. The second note just shows that the roots of $(x+1)^{2}-2$ are $x=\pm\sqrt{2}-1$ because $x'=x+1$ and $x'=\pm\sqrt{2}$ are the roots of $(x')^{2}-2$ and you can show that $K_{1}=\mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{2}-1)$. Likewise for the second factor (where the splitting field is $K_{2}=\mathbb{Q}(\sqrt[3]{7},\zeta_{3})$ where $\zeta_{3}$ is a primitive third root of unity). Now just show that $$\mathbb{Q}(\sqrt[3]{7},\zeta_{3})\cap\mathbb{Q}(\sqrt{2})=\mathbb{Q},$$ and consider the compositum $L=K_{1}K_{2}=\mathbb{Q}(\sqrt[3]{7},\zeta_{3},\sqrt{2})$ as an extension over $\mathbb{Q}$. $L/\mathbb{Q}$ is Galois and we have the following isomorphism of Galois groups $$\operatorname{Gal}(L/\mathbb{Q})\stackrel{\simeq}{\longrightarrow}\operatorname{Gal}(K_{1}/\mathbb{Q})\times\operatorname{Gal}(K_{2}/\mathbb{Q}),$$ which you can use to deduce the answer.