3 questions on field extensions I am trying to figure out some things regarding field extensions and some questions have arisen on the way.
Let $a$ be a positive integer which doesn't have a rational $nth$ root:


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*Is the splitting field of $x^n−a$ always equal to the splitting field of $x^n+a$? If not, when is this the case?

*When is the $nth$ cyclotomic polynomial irreducible over $Q(\sqrt[n]{a})$? If $gcd(a,n)=1$ is the cyclotomic polynomial then always irrecucible?

*Letting $a=1$. If $n\geq 3$ is the splitting field of $x^n-1$ equal to the splitting field of $x^n+1$? 
Comment on 3: For example, the minimal polynomial of the roots of $x^3-1$ (the 3rd roots of unity) is $x^2+x+1$ and the minimal polynomial of the roots of $x^3+1$ is $x^2-x+1$. Since $x^2-x+1$ is the minimal polynomial of the 6th roots of unity we should have that the splitting field of $x^3-1$ is contained in the splittingfield of $x^3+1$. Since the two fields have the same degree over $\mathbb{Q}$ this would imply that they are the same. Is this true in general? 
 A: Powers of $2$ are sometimes easy to ignore. The splitting field of $x^2-2$ is very different from the splitting field of $x^2+2$, and the splitting fields of $x^4-1$ and $x^4+1$ are different as well.
Your question #2 is not so easily answered as the others, I think.
A: Let's assume we are working over $\mathbb{Q}$ for simplicity with $\overline{\mathbb{Q}}$ a fixed algebraic closure.
Then the splitting field of $x^n-a$ is the smallest extension of $\mathbb{Q}$ containing all the $n$th roots of $a$, which are $a^{1/n}, \zeta_n a^{1/n}, \ldots , \zeta_n^{n-1} a^{1/n}$, where $a^{1/n}$ is some $n$th root of $a$ and $zeta_n$ is a primitive $n$th root of unity. On the other hand, the splitting field of $x^n + a$ is the smallest extension of $\mathbb{Q}$ containing all the $n$th roots of $-a$. Notice that $(-a)^{1/n} = (-1)^{1/n} a^{1/n}$. The difference is an $n$th root of $-1$. When does $-1$ have an $n$th root? When $n$ is odd, $(-1)^n = -1$. Hence, when $n$ is odd, $\mathbb{Q}$ already has an $n$th root of $-1$ and thus we need add nothing more to the splitting field of $x^n-a$ to get a splitting field of $x^n + a$.
This fact is nicely summarized in the question "Which roots of unity lie in $\mathbb{Q}(\zeta_n)$?" When $n$ is odd, then $\mathbb{Q}(\zeta_n)$ contains all the $2n$th roots of unity, and when $n$ is even it contains just the $n$th roots of unity. This should answer your third question.
So, to completely answer your first question: the splitting fields of $x^n - a$ and $x^n + a$ are the same if $n$ is odd. When $n$ is even, the latter polynomial splits in a quadratic extension of the former polynomials splitting field. In particular, $K = \mathbb{Q}(a^{1/n},\zeta_n)$ is a splitting field of $x^n-a$ and $K(\sqrt{\zeta_n})$ is a splitting field of $x^n + a$.
