Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with x^3 for polynomials f (x) and g(x).

Given p(x) be a nonzero polynomial of degree less than 1992 having no non constant factor in common with ${\displaystyle x^3-x}$
Let ${\displaystyle \frac{d^{1992}}{{dx}^{1992}}\left(\frac{p\left(x\right)}{x^3-x}\right)=\frac{f\left(x\right)}{g\left(x\right)}}$, where $f(x)$ and $g(x)$ are polynomials.
To find the smallest possible degree of $f(x)$:
BY division algorithm we know that,
If $f(x)$ and $g(x)$ are any two polynomials with ${\displaystyle g\left(x\right)\ne 0}$, then ${\displaystyle f\left(x\right)=g\left(x\right).q\left(x\right)+r\left(x\right)}$, where $r(x)=0$ or deg $r(x)$ where $q(x)$ is the quotient and $r(x)$ is the remainder.
Hence p(x) can be written as
${\displaystyle p\left(x\right)=(x^3-x)q\left(x\right)+r\left(x\right)}$
Here, we can see that the degree of $r(x)$ is 2 and the degree of $q(x)$is less than 1989
Then,
${\displaystyle \frac{d^{1992}}{{dx}^{1992}}\left(\frac{p\left(x\right)}{x^3-x}\right)=\frac{d^{1992}}{{dx}^{1992}}\left(\frac{r\left(x\right)}{x^3-x}\right)}$
To find ${\displaystyle \frac{r\left(x\right)}{x^3-x}}$
Using partial fractions, we have
${\displaystyle \frac{r\left(x\right)}{x^3-x}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x+1}}$
Hence,
${\displaystyle \frac{d^{1992}}{{dx}^{1992}}\left(\frac{r\left(x\right)}{x^3-x}\right)=1992!(\frac{A}{{(x-1)}^{1993}}+\frac{B}{x^{1993}}+\frac{C}{{(x+1)}^{1993}}}$
${\displaystyle =1992!\left(\frac{A{x}^{1993}{(x+1)}^{1993}+B{(x-1)}^{1993}{(x+1)}^{1993}+C{(x-1)}^{1993}{x}^{1993}}{{(x-1)}^{1993}{x}^{1993}{(x+1)}^{1993}}\right)}$

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