$\ x^3\:+\:a\left(a+1\right)x^2\:+\:ax\:-\:a\left(a+b\right)\:-\:1\:=\:0 $ $$\ x^3\:+\:a\left(a+1\right)x^2\:+\:ax\:-\:a\left(a+b\right)\:-\:1\:=\:0 $$
For what values of$\ b$ does the equation have a root which is independent of a?
Tried the Horner's Method, but doesn't seem to work with this. Could I have some hints on how to get this done? Thank you. *the answer is $\ b=2$.

HINT: write
$$x^3-1+a(a+1)x^2+ax-a(a+b)=0$$ Setting $$x=1$$ we get
$$a(a+1)+a-a(a+b)=0$$ and this is $$a(2-b)=0$$
if $a=0$ we get $x^3-1=0$
