Let $f \in \mathbb Q [X]$ and not constant or of the form $(x-a)^n$. Suppose:
$f_1 := \frac{f}{gcd(f,D^2f)}$ and;
$f_2 := \frac{f_1}{gcd(f_1,Df_1)}$,
where $Df$ stands for the formal derivative.
Is it true that $gcd(f_2,Df_2)=gcd(f_2,D^2f_2)=1$
Let $f \in \mathbb Q [X]$ and not constant or of the form $(x-a)^n$. Suppose:
$f_1 := \frac{f}{gcd(f,D^2f)}$ and;
$f_2 := \frac{f_1}{gcd(f_1,Df_1)}$,
where $Df$ stands for the formal derivative.
Is it true that $gcd(f_2,Df_2)=gcd(f_2,D^2f_2)=1$
No, it is not true. Try $f(x) = x^2 + x^4 $
$f_1(x) = x^2 + x^4 $
$f_2(x) = x + x^3 $
${\rm gcd}(f_2, D^2 f_2) = x$