jmc

We can see from the graph that there are three vertical asymptotes at $x=-2,1,2$. It follows that the denominator of the equation is given by $x^3+A{x}^2+Bx+C=(x+2)(x-2)(x-1)=(x^2-4)(x-1)=x^3-x^2-4x+4$. Thus, $A+B+C=-1-4+4=\boxed{-1}$.