What is the Galois Group of $f(t)=t^6-2t^3-1$ over $K:=\mathbb{Q}$? I have shown that $f(t)$ is irreducible, though I could not find an easier way to do this than by showing factorization into one polynomial of degree $4$ and one of degree $2$, or into two of degree $3$ were both impossible - this seems clunky, so I am surprised if it is necessary.
I've considered the splitting field of $g(t)=t^2-2t-1$, which is an intermediate extension $M$ such that $Gal(M:K)=C_2$. Since $M$ and $L$ are both normal over $K$, we have $Gal(M:K)=Gal(L:K)/M^*$, where $M^*$ is the subgroup of all $M$-automorphisms; I expect this to be useful. I have a few ideas from here, e.g. if $\beta_1,\beta_2$ are the roots of $g(t)$ I believe $L$ is the splitting field of $(x^3-\beta_1)(x^3-\beta_2)$ over $M$, but I am not managing to link these up into an actual description of the Galois group. Is this the right direction?
Thank you in advance for reading through this!
 A: I'll highlight some main steps, and leave some of the details and justifications for you to verify.
$(1)$ Let $f(x):=x^6-2x^3-1$ and $g(t):=t^2-2t-1$, so $x^3=t$. Then the roots of $g(t)=t^2-2t-1$ are $1\pm\sqrt 2$ so that the roots of $f(x)$ are $\sqrt[3]{1\pm\sqrt 2},~\omega\sqrt[3]{1\pm\sqrt 2},~\omega^2\sqrt[3]{1\pm\sqrt 2}$ with $\omega:=e^{2\pi i/3}$.
$(2)$ So the splitting field of $f(x)$ over $\Bbb Q$ is $M:=\Bbb Q(\sqrt[3]{1+\sqrt 2},\omega)$ which is a degree $12$ extension over $\Bbb Q$ since $[\Bbb Q(\sqrt[3]{1+\sqrt 2}):\Bbb Q]=6$ by irreducibility of $f(x)$ and $[\Bbb Q(\omega):\Bbb Q]=2$. Note that we omit $\sqrt[3]{1-\sqrt 2}$ from the generators of $M$ since $\sqrt[3]{1+\sqrt 2}=-(\sqrt[3]{1-\sqrt 2})^{-1}$. For complete rigour, you will need to verify this degree computation and the claim in the previous sentence.
$(3)$ It follows that $|\operatorname{Gal}(f/\Bbb Q)|=12$ so the Galois group of $f(x)$ over $\Bbb Q$ is $D_6$ (dihedral of order $12$) or $C_6\times C_2$ up to isomorphism, as the Galois group is a transitive subgroup of $S_6$.
$(4)$ We can show that the Galois group is not abelian, and thus $\operatorname{Gal}(f/\Bbb Q)\cong D_6$. To show non-abelian, use the fundamental theorem and exhibit a non-normal subfield of $M$.
