If a and b are integers such that ${\mathrm{x}}^2–\mathrm{x}–1$ is
a factor of ${\mathrm{ax}}^3+{\mathrm{bx}}^2+1$ then $\mathrm{b} =$

$\begin{array}{l}If{x}^2-x-1isafactorofa{x}^3+b{x}^2+1then\\ a{x}^3+b{x}^2+1=(cx+d)(x^2-x-1)\\ =c{x}^3-c{x}^2-cx+d{x}^2-dx-d\\ =c{x}^3+(d-c)x^2-(d+c)x-d\\ comparingwith\\ a=c,b=d-c,d+c=0,d=-1\\ Ifd=-1,thenc=1\\ \therefore b=-1-1\Rightarrow -2\end{array}$