When considering the polynomials $x^{n^2} \pm (x-1)^{n^2}$ ( $n$ integer > 1 ) i noticed some things that appeared weird to me.
Discriminant($x^{n^2} + (x-1)^{n^2}) = (n^2)^{n^2}$. Discriminant($x^{n^2} - (x-1)^{n^2}) = (n^2)^{n^2 -2}$.
But that was not all , it appears many of the roots of $x^{n^2} \pm (x-1)^{n^2} = 0$ apart from the trivial $\frac{1}{2}$ can be expressed in rootform ( for all $n$ ).
I just did the following but it seems i missed something :
$x^{n^2} \pm (x-1)^{n^2} = 0$
divide by $x^{n^2}$
$1 \pm (1-1/x)^{n^2} = 0$
$(1-1/x) = (\mp 1)^{1/n^2}$
abs formule
But that did not explain the rootforms or discriminant. And it surprised me those zero's could be expressed without the $(\mp 1)^{1/n^2}$ term.
I must say im not an expert in Discriminants or Galois theory. How to explain and prove this ?