Having trouble using my usual method of partial fraction decomposition for $\frac{9 + 3s}{s^3 + 2s^2 - s - 2}$.
You've written:
$$9+3s = A(s+1)(s+2)+B(s-1)(s+2)+C(s+1)(s-1).$$
This equation has to be true for any $s$, so choose clever values. For instance, $s=-1$ gives us
$$ 6 = B(-2)(1) \implies B=-3$$
Next $s=1$:
$$12 = A(2)(3) \implies A=2. $$
Next, $s=-2$:
$$3 = C(-1)(-3) \implies C=1.$$
One can also plug in complex values for $s$ and equate real and imaginary parts.
You've got the right expression $$ 9 + 3s = A(s + 1)(s + 2) + B(s - 1)(s + 2) + C(s + 1)(s - 1). $$ Now you can think like this: it is a polynomial equation of the kind $p(s)=0$. A nonzero polynomial cannot have more than the finite number of zeros. Since this equation has infinitely many solutions (all $s$ except a finite number) then the polynomial must be identical zero, and the equation holds true for those exceptional values as well. Setting $s=\pm 1, -2$ we get $A=2$, $B=-3$, $C=1$.
It is also possible - as you did - to set other values of $s$ and get the linear system. In your case you have obtained $$ \begin{cases} 2A-2B-C=9,\\ 12A+4B+3C=15,\\ 20A+10B+8C=18. \end{cases} $$ Using the linear algebra technique such systems are usually solved by Gauss elimination, but it is also possible to solve it by substitution as you tried, but got infinite loops. You have to do a systematic substitution: express one variable e.g. $C$ from one equation and substitute it into all other equations. Here how it goes: the first equation gives $C=2A-2B-9$. Substitute it into the second and the third ones $$ \begin{cases} C=2A-2B-9,\\ 12A+4B+3(2A-2B-9)=15,\\ 20A+10B+8(2A-2B-9)=18. \end{cases}\quad\Leftrightarrow\quad \begin{cases} C=2A-2B-9,\\ 9A-B=21,\\ 18A-3B=45. \end{cases} $$ Now we do the same: express $B$ from the second equation $B=9A-21$ and set into the third one $$ \begin{cases} C=2A-2B-9,\\ B=9A-21,\\ 18A-3(9A-21)=45. \end{cases}\quad\Leftrightarrow\quad \begin{cases} C=2A-2B-9,\\ B=9A-21,\\ A=2. \end{cases} $$ It works without infinite loops.