Solve the equation: $x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$ Here is the equation:

$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}$$

The equation looks very simple.  But, it's a little misleading (for me).
$$x-a^2x-\frac{b^2}{b^2-x^2}+a=\frac{x^2}{x^2-b^2}\Longrightarrow x(x^2-b^2)-a^2x(x^2-b^2)+b^2+a(x^2-b^2)-x^2=0\Longrightarrow (x^2-b^2)(x-a^2x+a-1)=0$$
We have  $x(1-a^2)=1-a$ and $x≠±b$.

1) $a=1$, $x\in (-\infty;b) ∪(-b;b)∪(b;\infty)$
2) $a=-1$ then $x\in\emptyset$
3) $a≠±1$ and $b≠±\frac{1}{a+1}$ ,then $x=\frac{1}{a+1}$

What I want to know, is there a mistake in the solution, or is there any other point I missed?
Thank you.
 A: Note that
$$\frac{b^2}{b^2-x^2}=-\frac{b^2}{x^2-b^2},$$
so, the equation becomes
$$x-a^2x+\frac{b^2}{x^2-b^2}+a=\frac{x^2}{x^2-b^2}$$
Passing $\frac{b^2}{x^2-b^2}$ to the RHS you obtain
$$x-a^2x+a=1.$$
This equation becomes $x(1-a^2)+a-1=0$.
The LHS becomes, $x(1-a)(1+a)-(1-a)=(1-a)\Big(x(1+a)-1\Big).$
So, if $a=1$, you have $x\in(-\infty,-b)\cup(-b,b)\cup(b,+\infty)$.
If $a=-1$ the equation has no solutions (it becomes 2=0).
If $a\neq-1$ and $a\neq1$, then $x=\frac{1}{1+a}.$
Now, in those solutions, we need to exclude when $x=b$ or $x=-b$. If $a=1$ we have already done that, so we just need to update it when $a\neq1$. In those cases, $x=\frac{1}{1+a}$, so if $\frac{1}{1+a}=\pm b$ the equation has no solutions.
To sum up,

If $a=-1$ the equation has no solutions.
If $a=1$,
  $x\in(-\infty,-b)\cup(-b,b)\cup(b,+\infty)$.
If $a\neq1$ and $\frac{1}{1+a}=\pm b$ the equation has no solutions.
Otherwise, $x=\frac{1}{1+a}$.

A: Your solution is almost right, except that 2) is not a solution because that yields $0=2$ in your equation.
Here’s a simpler method.
Adding $\frac{b^2}{b^2-x^2}$ to both sides, note that the RHS now becomes $1$. So we now need to solve $x(1-a^2)=1-a$, which factorises to $(1-a)(x(a+1)-1)=0$ (where $x \neq \pm b$).
