Hungerford’s book of algebra has exercise 6 chapter 3 section 6 [Probably impossible with the tools at hand.]:

Let p∈Z be a prime; let F be a field and let c∈F. Then xp−c is irreducible in F[x] if and only if xp−c

has no root in F. [Hint: consider two cases: char(F)=p and

char(F) different of p.]I have attempted this a lot. Anyone has an answer?

**Answer**

Perhaps the simplest tool I can think of is the following:

Let F be a field and f(x) an irreducible polynomial over F, then there is a field K≥F where f(x) has a root; as f(x) is irreducible in F[x], a principal ideal domain, then ⟨f(x)⟩ is a maximal ideal of F[x], hence K=F[x]/⟨f(x)⟩ is a field, ˉx is a root of f(x), and it is easy to see how to embed F into K. Now given a polynomial f(x)∈F[x] it is clear how to construct a field K≥F such that f(x) factors into linear polynomials in K[x].

Now your question can be answered as follows:

Let K≥F be a field where xp−c factors into linear polynomials, say xp−c=(x−z1)⋯(x−zp). Suppose xp−c is not irreducible in F[x], then there are polynomials f(x),g(x)∈F[x] of degree ≥1 such that xp−c=f(x)g(x), then we may assume f(x)=(x−z1)⋯(x−zn), where degf(x)=n<p.

Put z=z1⋯zn, then z is ± the constant term of f(x), so z∈F, and zp=(z1⋯zn)p=zp1⋯zpn=cn. As p is prime there are integers a,b such that 1=ap+bn, then (cazb)p=capzbp=capcbn=c,

but cazb∈F, so xp−c has a root in F.

**Attribution***Source : Link , Question Author : user79709 , Answer Author : Camilo Arosemena-Serrato*