Find the degree of $\mathbb{Q}(i,\sqrt[4]{3}, \sqrt[6]{3})$ over $\mathbb{Q}$ 
Find the degree of $\mathbb{Q}(i,\sqrt[4]{3}, \sqrt[6]{3})$ over
  $\mathbb{Q}$

Since $\mathbb{Q}(i)$ is contained in either $\mathbb{Q}$ of the roots of $x^4-3$ or $x^6-3$, we can just take the degree of $\mathbb{Q}(\sqrt[4]{3}, \sqrt[6]{3})$ over $\mathbb{Q}$, right? Then, by the multiplicative formula of the degrees, it's gonna be $4\cdot 6 = 24$. Am I right?
 A: $$\begin{array}{rcl}
|\Bbb Q(i,\sqrt[4]3,\sqrt[6]3):\Bbb Q|
&=& |\Bbb Q(i,\sqrt[4]3,\sqrt[6]3):\Bbb Q(\sqrt[4]3,\sqrt[6]3)| ~ |\Bbb Q(\sqrt[4]3,\sqrt[6]3):\Bbb Q| \\
&=& 2|\Bbb Q(\sqrt[4]3,\sqrt[6]3):\Bbb Q| \\
&=& 2|\Bbb Q(\sqrt[4]3,\sqrt[6]3):\Bbb Q(\sqrt[4]3,\sqrt3)| ~ |\Bbb Q(\sqrt[4]3,\sqrt3):\Bbb Q| \\
&=& 2 \times 3 \times |\Bbb Q(\sqrt[4]3,\sqrt3):\Bbb Q| \\
&=& 2 \times 3 \times |\Bbb Q(\sqrt[4]3):\Bbb Q| \\
&=& 2 \times 3 \times 4 \\
&=& 24
\end{array}$$
As various people have pointed out, $\Bbb Q(i,\sqrt[4]3,\sqrt[6]3) = \Bbb Q(i,\sqrt[12]3)$, so its Galois group over $\Bbb Q$ contains a $12$-cycle and an involution. $\Bbb Q(\sqrt[12]3)$ is not normal, so the Galois group is not abelian, so it is $D_{24}$ (or $D_{12}$ depending on convention), the dihedral group of order $24$.
A: After using 
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc);$$
$$2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4=(a+b+c)(a+b-c)(a+c-b)(b+c-a)$$ and
$$(a-bi)(a+bi)=a^2+b^2$$ we'll get a polynomial from $\mathbb Q[x]$ with degree $3\cdot4\cdot2=24$.
