Splitting Field Problem I have a problem on splitting field as follows:
Determine the degree of the splitting field of the polynomial $x^6-7$ over:
(a) $\mathbb{Q},$
(b) $\mathbb{Q(\alpha)},$ where $\alpha$ is primitive 3rd root of unity,
(c) $\mathbb{F_3},$ (field with 3 elements).
I try to prove (a) in the following way:
Since $x^6-7=(x^3-\sqrt7)(x^3+\sqrt7),$ then the splitting field $K$ for $x^6-7$  must contain the splitting fields for $(x^3-\sqrt7)$ and $(x^3+\sqrt7).$ The roots of $(x^3-\sqrt7)$ are $7^{1/6}, 7^{1/6}\xi, 7^{1/6}\xi^2,$ and the roots of $(x^3+\sqrt7)$ are $-7^{1/6}, -7^{1/6}\xi, -7^{1/6}\xi^2,$ where $\xi$ is a primitive sixth root of unity.  Is this reasoning correct?
How do I solve parts (b) and (c)? Thanks in advance!

As $\sqrt[6]{7}$ is a zero of the polynomial and $\sqrt[6]{7} \not \in \mathbb{Q}$ we adjoin it to $\mathbb{Q}$ to obtain $\mathbb{Q}(\sqrt[6]{7})$. Now in $\mathbb{Q}(\sqrt[6]{7})$ the polynomial factors into $(x - \sqrt[6]{7})(x + \sqrt[6]{7})(x^2 + \sqrt[6]{7}x + \sqrt[3]{7})(x^2 - \sqrt[6]{7}x + \sqrt[3]{7})$. Now the roots of the quadratic factors are $\xi\sqrt[6]{7},\xi^2\sqrt[6]{7},-\xi\sqrt[6]{7},-\xi^2\sqrt[6]{7}$, where $\xi$ is the third root of unity. As $\xi \not \in \mathbb{Q}(\sqrt[6]{7})$. Adjoining we obtain that $\mathbb{Q}(\sqrt[6]{7}, \xi)$ is the splitting field of the polynomial over $\mathbb{Q}$. Now we have:
$$[\mathbb{Q}(\sqrt[6]{7}, \xi) : \mathbb{Q}]=[\mathbb{Q}(\sqrt[6]{7},\xi) : \mathbb{Q}(\sqrt[6]{7})][\mathbb{Q}(\sqrt[6]{7}) : \mathbb{Q}] = 2 \cdot 6 = 12$$
This is true as the minimal polynomial of $\xi$ over $\mathbb{Q}(\sqrt[6]{7})$ is $x^2 + x + 1$ and the minimal polynomial of $\sqrt[6]{7}$ over $\mathbb{Q}$ is $x^6 - 7$. Therefore the degree is $12$.
For the second part as $\mathbb{Q}(\xi) = \mathbb{Q}(\xi^2)$ and using the previous part we have that the splitting field is again $\mathbb{Q}(\sqrt[6]{7}, \xi)$, but this time the degree is $6$.
For the third part we have that $x^6 - 7 = (x+1)^3(x+2)^3$ in $\mathbb{F}_3$, so the splitting field is infact $\mathbb{F}_3$ and the degree is of order $1$.
