Find the degree of $[F:\mathbb{Q}]$, where $F$ is the splitting field of $x^3-11$ over $\mathbb{Q}$ 
Find the degree of $[F:\mathbb{Q}]$, where $F$ is the splitting field
  of $x^3-11$ over $\mathbb{Q}$

Roots are $\sqrt[3]{11}$ and the other $2$ are complex, I think. I don't think that finding the roots and then what field they generate is a good idea. I think it's better to prove this polynomial is irreducible. 
By Eiseintein criterion, $p=11$ is prime, does divide every $a_i$ except $a_3=1$, $p^2$ does not divide $a_0$. So the polynomial is irreducible over $\mathbb{Q}$ so the degree is $3$. Is this right?
 A: There is a more general result: 

For all prime numbers $p$ the splitting field $F$ of $f=x^3-p$ over $\mathbb{Q}$ has degree 6.

Reason: The polynomial is irreducible by Eisenstein criterion and has exactly one real root $z_1=\sqrt[3]{p}$, because the polynomial is strictly increasing. Now there must be a pair of non-real roots $z_2$ and $z_3$. Because the coefficients of $f$ are real, we have $\overline{z_2}=z_3$.
Now I use the following theorem

If $f$ is an irreducible polynomial of degree $n$ over $\mathbb{Q}$, there is a splitting field $F$ of $f$ with $[F:\mathbb{Q}]\leq n!$

Let $F$ be the splitting field of $x^3-p$. Remark that $F=\mathbb{Q}(z_1,z_2,z_3)$ is a splitting field of $f$ over $\mathbb{Q}$ up to isomorphism. The theorem says $[F:\mathbb{Q}]\leq 3!=6$. Note that $[\mathbb{Q}(z_1):\mathbb{Q}]=3$, because $f$ is the minimal polynomial of $z_1$. But both $z_2,z_3\notin\mathbb{Q}(z_1)$, thus $1<[\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}(z_1)]\leq 2$ and consequently 
\begin{align}
[\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}]=[\mathbb{Q}(z_1,z_2,z_3):\mathbb{Q}(z_1)]\cdot [\mathbb{Q}(z_1):\mathbb{Q}]=2\cdot 3=6
\end{align}
If I am missing some details or there is a mistake, please let me know.
A: HINT: The roots are:
 $$a_1 = \sqrt[3]{11} \quad \quad a_2  = \xi_{3}\sqrt[3]{11} \quad  a_2  = \xi^2_{3}\sqrt[3]{11}$$
where $\xi_{3}$ is the third root of unity. Now you need an extension of $\mathbb{Q}$ that contains both $\xi_{3}$ and $\sqrt[3]{11}$
A: The roots of $x^3-11$ are $11^{\frac{1}{3}}$ ,$11^{\frac{1}{3}}\omega$ ,$11^{\frac{1}{3}} \omega^2$
Hence $F=\Bbb Q(11^{\frac{1}{3}},w)$
Hence $[F:\Bbb Q]=[\Bbb Q(11^{\frac{1}{3}},w):\Bbb Q(11^{\frac{1}{3}})][\Bbb Q(11^{\frac{1}{3}}):\Bbb Q]=2\times 3=6$
[Since the minimal polynomial of $\omega$ is $x^2+x+1$ and minimal polynomial of $11^{\frac{1}{3}}$ over $\Bbb Q$ is $x^3-11$.]
A: Your argument that $x^3 - 11$ is irreducible over $\Bbb Q$ is correct, but this does not imply that the degree of the extension is $3$. (It does, however, imply that the degree is at least $3$; we can thus conclude that the only possible degrees of a splitting field of an irreducible cubic are $3$ and $6$.)
Hint Since the polynomial has nonreal roots, (the restriction to $F$ of) complex conjugation is an automorphism of $F / K$ of order $2$.
