$\ 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 $\ b=2$.

Put x = 1
You will find that for b = 2, irrespective of the value of a, you will find a root.  It is that simple.  But not scientific.
$1+a(a+1) + a - a(a+b) -1 = 2a-ab = 0$
In the above expresssion, if you put b=2, the root does not have to be dependent on a
