jmc

The $x+5$ and $(x-3{)}^2$ in the denominators suggest that these might be factors of $x^3-x^2-21x+45$. Indeed, we find that this polynomial equals $(x+5)(x-3{)}^2$. Clearing denominators, we find that $$1=A(x-3{)}^2+B(x+5)(x-3)+C(x+5).$$Thus, when we substitute $x=-5$, we find that $(-5-3{)}^{2}A=64A=1$, so $A=\boxed{\frac{1}{64}}$.