Special representation of polynomial How to prove that for natural $n$ the polynomial $(x^4-6x^2+1)^n$ can't be represented in such a way 
 $$
 (x^4-6x^2+1)^n=f(x)^2+1,  (x^4-6x^2+1)^n=g(x)^3-1, 
 $$
  where$f(x), g(x)$  are polynomials.
Seems, for  $n=2k$ I can prove it for the first  case. We have
  $$
  1=f(x)^2-(x^4-6x^2+1)^{2k}=(f(x)-(x^4-6x^2+1)^{k})(f(x)+(x^4-6x^2+1)^{k}).
  $$
  Thus $f(x)-(x^4-6x^2+1)^{k}=c, f(x)+(x^4-6x^2+1)^{k}=c^{-1}$ or $f(x)-(x^4-6x^2+1)^{k}=-c, f(x)+(x^4-6x^2+1)^{k}=-c^{-1}$ for some constant $c.$  It is easy to show that this leads to a contradiction.
How it may be proved for  case $n=2k+1$ and for the second case $(x^4-6x^2+1)^n=g(x)^3-1.$?

Suppose $F(x)=(x^4-6x^2+1)^n-1$ is the square of some polynomial for some positive integer $n$. Then $F(x)$ has a zero at $x_0$ only if $F'(x_0)=0$ as well. But $F(x)=0$ for $x=\pm \sqrt{6}$, whereas $$F'(\pm \sqrt{6})=\bigg[4nx(x^2-3)(x^4-6x^2+1)^{n-1}\bigg]_{x=\pm\sqrt{6}}=\pm 12n\sqrt{6}\neq 0$$
for any positive integer $n$. Hence $F(x)$ is not the square of a polynomial.
Can you generalize this to the cubic case?
