Show that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})$ is a splitting field for $f(x)$. a) Show that $f(x)=x^6-2x^3-10$ is irreducible over $\mathbb{Q}$.
Proof: Using Eisenstein's Criterion, with $p=2$, we show $f(x)$ is irreducible over $\mathbb{Q}$.
b) Let $\alpha=\sqrt[3]{1+\sqrt{11}}$. Show that $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.
Proof: Note that $\alpha$ is a root of $f(x)$. By part (a), we can conclude that $f(x)$ is the minimal polynomial of $\mathbb{Q}(\alpha)$. Furthermore $|\mathbb{Q}(\alpha):\mathbb{Q}|=6$.
c) Show that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})$ is a splitting field for $f(x)$.
Proof: Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.]
$$ \alpha \cdot \beta = \sqrt[3]{(1-\sqrt{11})(1-\sqrt{11})} = \sqrt[3]{-10} $$

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*How do I get rid of the negative in my radical? Recall $\sqrt[3]{-1}=-1$.

*What combination will help me get $\sqrt{-3}$? Refer to: Determine splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$
d) Show that $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$ and conclude that $|L:\mathbb{Q}|=12$, where $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. Moreover $K=L(\alpha)$ and thus $|K:\mathbb{Q}|=12$ or $36$.

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*How do I show $\sqrt[3]{10}\not\in \mathbb{Q}(\sqrt{-3},\sqrt{11})$?

*Since the corresponding minimal polynomials of $\sqrt{-3},\sqrt{11},\sqrt[3]{10}$ have degree $2,2,3$, respectively, and share no roots, $|L:\mathbb{Q}|=12$.

*$L(\alpha)= \mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})(\alpha)$. Does $\alpha$ get consumed by $\sqrt[3]{10}$ from what we shown in part (c) and the first part in (d)? Is this how we get $K=L(\alpha)$?

*How can $|K:\mathbb{Q}|=36$?

 A: Answers for (a) and (b) are in the post.
Proof of (c): Note that $\beta=\sqrt[3]{1-\sqrt{11}}$ is also a root of $f(x)$. [Check by plugging it in $f(x)$.]
$$\alpha\cdot\beta= \sqrt[3]{(1+\sqrt{11})(1-\sqrt{11})}=\sqrt[3]{-10}=-\sqrt[3]{10}$$
Since $\beta\in\mathbb{Q}(\alpha)$, $-\alpha\beta\in\mathbb{Q}(\alpha)$. Hence $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\sqrt[3]{10})$.
Recall the idea of roots of unity. The roots of $f(x)$ are $x=\alpha, \omega\alpha, \omega^2\alpha$ where $\omega=\dfrac{-1+i\sqrt{3}}{2}$. It follows that $K=\mathbb{Q}(\alpha,\sqrt{-3},\sqrt[3]{10})=\mathbb{Q}(\alpha,\sqrt[3]{10})$. Furthermore $K$ is a splitting field for $f(x)$.
Proof of (d): Since $|\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}|=3$, and $|\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}|=4$, $\sqrt[3]{10}\not\in\mathbb{Q}(\sqrt{-3},\sqrt{11})$. (why??) I know we said Tower Law, but I think I need a review of it.
Let $L=\mathbb{Q}(\sqrt{-3},\sqrt{11},\sqrt[3]{10})$. By Tower Law,
$$ |L:\mathbb{Q}|=|L:\mathbb{Q}(\sqrt{-3},\sqrt{11}):\mathbb{Q}(\sqrt[3]{10})|\cdot |\mathbb{Q}(\sqrt[3]{10}):\mathbb{Q}| = 4\cdot 3 =12$$
Let $K=L(\alpha)$, if $\alpha$ is real, then just like the last part, we have
$$|K:\mathbb{Q}|=12$$
Recall the  $\alpha$ is a root of unity that, then we have 2 more roots to account more, so
$$|K:\mathbb{Q}|=12\cdot 3 = 36.$$
Comments on the proofs will be greatly appreciated.
A: Supplementing the other answer with a possibly unnecessarily high-browed argument proving that the splitting field has degree $36$. Or, equivalently, that $1+\sqrt{11}$ has no cube root in the field $L$.
The tool I use is Dedekind's theorem, so the argument relies on Galois theory and group actions. Consider the polynomial $f(x)=x^6-2x^3-10$. Let $G$ be the Galois group of this polynomial, viewed as a subgroup of the group of permutations of the six roots, identified with $S_6$.

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*Eisenstein's criterion ($p=2$) tells us that $f(x)$ is irreducible over $\Bbb{Q}$. Therefore $G$ acts transitively on the set of roots, and hence
$6\mid |G|$.

Let us fix a root $\alpha$ of $f$ in the splitting field $K$ (unique as a subfield of $\Bbb{C}$). Factorization of $f$ modulo different primes reveals other things about the action of $G$. I will pay particular attention to the subgroup $H=Stab_G(\alpha)$ that keeps the root $\alpha$ fixed.

*

*Modulo $7$ we have $f(x)=(x^3+1)(x^3+4)$. The first cubic factors fully, $x^3+1=(x+1)(x+2)(x+4)$, as the field $\Bbb{F}_7$ has three third roots of unity: $1,2,4$. On the other hand, the cubic $x^3+4$ remains irreducible modulo $7$ because the congruence $x^3+4\equiv0\pmod7$ has no integer solutions. So Dedekind tells us that $G$ contains a $3$-cycle $\sigma$. This $3$-cycle fixes three of the zeros. Because $G$ acts transitively on the set of roots, one of the conjugates of $\sigma$ in $G$ has $\alpha$ as a fixed point. It follows that $H$ contains a $3$-cycle, and hence $3\mid |H|$.

*By the orbit-stabilizer theorem $|G|=6|H|$ is thus a multiple of $18$. The other answer shows that $|G|=[K:\Bbb{Q}]\in\{12,36\}$, so we can conclude that
$$|G|=[K:\Bbb{Q}]=36.$$

This was somewhat unsatisfactory in the sense that initially I needed help from a computer program. At some point this becomes necessary as eliminating alternative Galois groups becomes more and more demanding, when the degree grows (barring some special cases).
In the thread inspired by this Paramanand Singh reaches the same conclusion by a different computer aided calculation proving that an element of $K$ has a minimal polynomial of degree nine, implying $9\mid |G|$ and again eliminating $[K:\Bbb{Q}]=12$ as an alternative.
