<s>[INST] Partial fractions of $\frac{x^4}{(x^2-1)^3}$ (is it possible to find coefficients without expanding?) I want to decompose the fraction $\frac{x^4}{(x^2-1)^3}$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac D{x+1}+\frac E{(x+1)^2}+\frac F{(x+1)^3}$$
Because $f(x)=\frac{x^4}{(x^2-1)^3}$ is an even function we have $f(x)=f(-x)$. Hence we have $A=-D\;,\;B=E\;,\;C=-F$ :
$$\frac{x^4}{(x^2-1)^3}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{(x-1)^3}+\frac {-A}{x+1}+\frac B{(x+1)^2}+\frac {-C}{(x+1)^3}$$
Hence:
$$A(x-1)^2(x+1)^3+B(x-1)(x+1)^3+C(x+1)^3-A(x+1)^2(x-1)^3+B(x+1)(x-1)^3-C(x-1)^3=x^4$$
If we plug in $x=1$ we get $C=\frac18$. but I can't find $A$ and $B$ without expanding $(x-1)^2(x+1)^3\ldots$ is it possible to find them without expanding whole expression?
 [/INST] Following my comments, we may try two simple values of $x$ to reduce computations. Here I believe $x=0$ and $x=2$ are simple enough.
Plugging in $x=0$ gives $$A-B+C-(-A)+(-B)-(-C)=0\implies A-B=-\frac 1 8$$
and plugging in $x=2$ gives $$A\cdot 3^3 - B\cdot 3^3 + C\cdot 3^3 - A\cdot 3^2 + B\cdot 3 - C = 2^4\implies A + \frac 5 3 B =\frac {17} {24}$$
which gives us $A=\frac 3 {16}, B=\frac 5 {16}$.
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