Find constants of function I have this equality : 
$$f(x)=\frac{9}{(x-1)(x+2)^2}$$
I am required to find the constants A, B and C so that,
$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}} $$
How do we go about solving such a question? 
I am not sure on how to solve such questions. Approach and Hints to solve these kinds of questions are welcomed. :)
Thank you!
Edit:
Wow, saw all the answers! Didn't know that there were a lot of different ways to solve this question. Math is such a fascinating thing!

$$\dfrac{9}{(x-1)(x+2)^2}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$$
Multiplying by the GCD(which means multiplying all the terms by $(x-1)(x+2)^2$)
$$9=A(x+2)^2+B(x-1)(x+2)+C(x-1)$$
Let $x=-2$;
$$9=A(-2+2)^2+B(-2-1)(-2+2)+C(-2-1)$$
$$9=A(0)+B(-3)(0)+C(-3)$$
$$9=-3C$$
$$C=-3$$
Now that we have C,we let x=1;
$$9=A(1+2)^2+B(1-1)(1+2)+C(1-1)$$
$$9=A(3)^2+B(0)(3)+C(0)$$
$$9=9A$$
$$\implies A=1$$
then we let x=3 and substitute back A and C
$$9=(3+2)^2+B(2)(5)-3(2)$$
$$9=25+10B-5$$
$$9=20+10B$$
$$-11=10B$$
$$B=\dfrac{-11}{10}$$
So,now you have the values of A,B and C. This is how you go about doing it. Try it yourself. There might be a small arithmetic error in this solution. Solve it yourself to find it.
