Let $A$, $B$, and $C$ be constants so that \[\frac{1}{x^3-x^2-21x+45}=\frac{A}{x+5}+\frac{B}{x-3} + \frac{C}{(x - 3)^2}\]holds for all real numbers $x$ other than $-5$ and $3.$ What is $A$?
x^3 - x^2 -21x + 45 factors as (x-3)^2 (x + 5)
Multiply through by this factorization and we get
1 = A(x-3)^2 + B (x-3) ( x + 5) + C(x + 5)
Let x = -5
Then A(-8)^2 = 1 → A = 1/64
+28 
