$\ 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 $\ b=2$.
 A: Why not rearrange to give $$ab=x^3+a(a+1)x^2+ax-a^2-1$$
Divide through by $a$ to obtain: $$b=a(x^2-1)+\frac {x^3-1}a + x^2+x$$
Now you can set $x$ as a root of the equation - when does this lead to an expression independent of $a$?
A: 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
A: 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 $x^3-1=0$
A: If the root is independent of $a$ then the function mustn't be changing with respect to $a$ at that point so look at the derivative with respect to $a$:
$$\frac{d}{da}x^3+a(a+1)x^2+ax-a(a+b)-1=(2a+1)x^2+x-2a-b$$
Solving this for $x$ gives solutions of:
$$x=\frac{-1\pm\sqrt{1+8a+16a^2+4b+8ab}}{2(1+2a)}$$
For the root to be independent of $a$ and as the expression under the root has a $16a^2$ then $x$ must be 1.
(Other answers can then follow from here.)
