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.

use $$(1-x)^{-1}=1+x+x^2 +x^3...$$
and $$(1+x)^{-1}=1-x+x^2-x^3....$$
your question is of the form
$$3^{-1}.(-2)^{-2}(\frac{x}{3}+1)^{-1}(1-\frac{x}{2})^{-2}$$
generally
$$(1-x)^{-k}=1+kx+\frac{k(k+1)}{2!}x^2+ \frac{k(k+1)(k+2)}{3!}x^3+.... $$
now to find coefficient of $x^8$ . group all possible terms so that the resulting degree of the term is $8$.
i think you should get $0.002706025...$
