jmc

If $a{x}^3+bx-c$ has $x^2+bx+c$ has a factor, then the other factor must be of the form $ax-1,$ in order to make the leading and constant coefficients match. Thus, $$(x^2+bx+c)(ax-1)=a{x}^3+bx-c.$$Expanding, we get $$a{x}^3+(ab-1)x^2+(ac-b)x-c=a{x}^3+bx-c.$$Matching coefficients, we get $$\begin{array}{rl}ab-1& =0,\\ ac-b& =b.\end{array}$$Hence, $ab=\boxed{1}.$