jmc

If $x^2-x-1$ is a factor of $a{x}^3+b{x}^2+1,$ then the other factor must be linear, where the coefficient of $x$ is $a,$ and the constant coefficient is $-1.$ Thus $$(x^2-x-1)(ax-1)=a{x}^3+b{x}^2+1.$$Expanding, we get $$a{x}^3-(a+1)x^2+(1-a)x+1=a{x}^3+b{x}^2+1.$$Matching coefficients, we get $$\begin{array}{rl}-(a+1)& =b,\\ 1-a& =0.\end{array}$$Hence, $a=1.$ Then $b=-(a+1)=\boxed{-2}.$