Let $p(x)$ be a polynomial of degree strictly less than 100 and such that it does not have $x^3-x$ as a factor. If $$\frac{d^{100}} {dx^{100}} \left(\frac{p(x)}{x^3-x}\right)=\frac{f(x)} {g(x)}$$ for some polynomials $f(x)$ and $g(x)$. Then, find the smallest possible degree of $f(x)$.
Here $$\frac{d^{100}}{dx^{100}}$$ means taking $100$th* derivative.
My attempt
$$\frac{p(x)}{x^3-x}=q(x) +\frac{r(x)} {x^3-x}$$ Then, if $p(x)$ has degree strictly less than 100 then it may be of 99 or 98 or something like that. But if it is like that then its 100th derivative will be zero.
As in case of $x^2$ its first derivative is $2x$, then its second derivative is 2 and then the third derivative is 0. So, if f(x) is nth degree polynomial its n+1 derivative is 0. But then it does not satisfies the given conditions.
Edit: The test solution gives another answer of 200 with another solution giving 0 as answer. Now how can two least possible degrees can be the solution
Long division and partial fraction decomposition produce $${p(x)\over x^3-x}=q(x)+{A\over x}+{B\over x-1}+{C\over x+1}\ ,$$ with $q$ a polynomial of degree $\leq96$ and $A$, $B$, $C$ constants, not all of them $=0$. Now $A=1$, $B=C=0$ is possible. In this case $${d^{100}\over dx^{100}}\left({p(x)\over x^3-x}\right)={100!\over x^{101}}\ .$$ The answer to the question therefore is $0$.