Is $\sqrt[3]5+\sqrt[3]7$ contained in a cyclotomic extension? Is $\sqrt[3]5+\sqrt[3]7$ contained in a cyclotomic extension?
I am trying to prove that $\Bbb Q(\sqrt[3]5+\sqrt[3]7)/ \Bbb Q $ is an abelian or non abelian extension, so I will prove that this is either a cyclotomic extension or not correspondingly.
But I can't even find the minimal polynomial for this extension 
and in particular to find the roots of this polynomial.
By looking at the appropriate authomorphisms acting on those roots I will learn about its abelian (or non abelian) property.  
 A: Here's how you can prove that $\mathbb{Q}(\sqrt[3]{5} + \sqrt[3]{7}) = \mathbb{Q}(\sqrt[3]{5},\sqrt[3]{7})$.
Let $f(x) = x^3 - 5$. Now consider the polynomial $f((\sqrt[3]{5}+\sqrt[3]{7})-x) \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})[x]$. Obviously the roots of it are $\sqrt[3]{7}$, $\sqrt[3]{5}+\sqrt[3]{7}-\zeta_3\sqrt[3]{5}$, $\sqrt[3]{5} + \sqrt[3]{7} - \zeta_3^2\sqrt[3]{5}$, where $\zeta_3$ is the third root of unity.
On the other hand $\sqrt[3]{7}$ is a root of $x^3 - 7 \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})[x]$, whose roots are $\sqrt[3]{7},\zeta_3\sqrt[3]{7},\zeta_3^2\sqrt[3]{7}$. 
Now the minimal polynomial of $\sqrt[3]{7}$ divides both of the polynomials above and in particular we must have that it divides $x-\sqrt[3]{7}$, as $\sqrt[3]{7}$ is the only common root of the polynomials. Obviously the minimal polynomial has to have a positive degree and so we conclude that it's $x-\sqrt[3]{7}$ itself. Therefore $x-\sqrt[3]{7} \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})[x]$ and in particular $\sqrt[3]{7} \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$. From here it's easy to conclude that $\sqrt[3]{5} \in \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$ and so $\mathbb{Q}(\sqrt[3]{5},\sqrt[3]{7}) \subseteq \mathbb{Q}(\sqrt[3]{5}+\sqrt[3]{7})$. The other inclusion is trivial and hence the claim is proven.
Now you can see that $\mathbb{Q}(\sqrt[3]{5},\sqrt[3]{7},\zeta_3)$ is the Galois closure of the field above, which isn't an abelian extension and so $\sqrt[3]{5}+\sqrt[3]{7}$ isn't contained in any cyclotomic extension.

[UPDATE]: Here's another and easier solution to the problem. Assume that $\sqrt[3]{5} + \sqrt[3]{7}$ is contained in some cyclotomic extension. Then as the Galois group over $\mathbb{Q}$ of the later is abelian we have that every intermediate subfield is normal. In particular we have that $\mathbb{Q}(\sqrt[3]{5} + \sqrt[3]{7})$ is normal over $\mathbb{Q}$. The minimal polynomial of $\sqrt[3]{5}+\sqrt[3]{7}$ over $\mathbb{Q}$ is $x^9-36x^6-513x^3-1728$ and obviously $\zeta_3(\sqrt[3]{5} + \sqrt[3]{7})$ is a root of it. However it's not an element of $\mathbb{Q}(\sqrt[3]{5} + \sqrt[3]{7})$, as the later is a real subfield. This contradicts the normality of $\mathbb{Q}(\sqrt[3]{5} + \sqrt[3]{7})$, which is the wanted contradiction.
