If the equations $x^2+bx-1=0$  and  $x^2+x+b=0$  have a common root different from – 1, then  ${\left|b\right|}^4$ is equal to

$x^2+bx-1=0$  [common root]
 $x^2+x+b=0$  $x=\frac{b+1}{b-1}$
 Put  $x=\frac{b+1}{b-1}$  in equation
 $$\begin{gathered} {\left(\frac{b+1}{b-1}\right)}^2+\left(\frac{b+1}{b-1}\right)+b=0 \\ \Rightarrow {(b+1)}^2+(b+1)(b-1)+b{(b-1)}^2=0 \\ \Rightarrow b^2+1+2b+b^2-1+b(b^2-2b+1)=0 \\ \Rightarrow 2b^2+2b+b^3-2b^2+b=0\Rightarrow b^3+3b=0 \\ \Rightarrow b(b^2+3)=0\Rightarrow b^2=-3 \\ \Rightarrow b=\pm \sqrt{3}i\Rightarrow {\left|b\right|}^4=9 \end{gathered}$$