<s>[INST] How to find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$ Using which way can we find coefficient of $x^8$ in $\frac{1}{(x+3)(x-2)^2}$? I have used binomial theorem but failed to find an answer for it.
 [/INST] You can decompose
$$
\frac{1}{(x+3)(x-2)^2}=\frac{A}{1+x/3}+\frac{B}{1-x/2}+\frac{C}{(1-x/2)^2}
$$
and it's just a matter of finding $A$, $B$ and $C$. For $x=0$, $x=1$ and $x=-1$, we get
$$
\begin{cases}
\frac{1}{12}=A+B+C\\
\frac{1}{4}=\frac{3}{4}A+2B+4C\\
\frac{1}{18}=\frac{3}{2}A+\frac{2}{3}B+\frac{4}{9}C
\end{cases}
$$
or
$$
\begin{cases}
A+B+C=1/12\\
3A+8B+16C=1\\
27A+12B+8C=1
\end{cases}
$$
This gives
$$
A=\frac{1}{75},\quad B=\frac{1}{50},\quad C=\frac{1}{20}
$$
(if I computed right).
Now you can recall that
$$
\frac{1}{1+x/3}=\sum_{k\ge0}(-1)^k\frac{x^k}{3^k},
\qquad
\frac{1}{1-x/2}=\sum_{k\ge0}\frac{x^k}{2^k},
$$
and that $1/(1-x/2)^2$ is proportional to the derivative of $1/(1-x/2)$, precisely
$$
\left(\frac{1}{1-x/2}\right)'=\frac{1}{2}\frac{1}{(1-x/2)^2}.
$$
Therefore
$$
\frac{1}{(1-x/2)^2}=2\sum_{k\ge0}(k+1)\frac{x^k}{2^k}
$$
and the coefficient you're looking for is
$$
A\frac{1}{3^8}+B\frac{1}{2^8}+2C\frac{9}{2^8}.
$$
</s>