I still maintain that my answer is right. any naysayers?

 

The qn only asks us to find any factor of f(xⁿ) not particularly to prove that (xⁿ-1) is a factor. You can easily see that (xⁿ-1)  is a factor automatically implies that (x-z) where z is a solution to zⁿ=1 becomes a factor.

 

@Saurabh:

Let f(x) = a_nxⁿ+a_n-1x^n-1+...+a₀

Then f(xⁿ) = a_n(xⁿ)ⁿ + a_n-1x^n-1+..+a₀.

f(1) = a_n+a_n-1+...+a₀ = 0 as x-1 is a solution

Now you consider any solution to zⁿ=1 and find f(xⁿ) where x = z you get

f(zⁿ) = f(1) = 0

 

f(xⁿ) = F(x) where F is a polynomial of degree n² (of which a_n is the coefficient). and F(z) = 0. Hence (x-z) is a factor of F(x) and hence of f(xⁿ).