Solve $(x^2+1)(x-1)^2=ax^2, \ a\in\mathbb{R}$ 
Solve $(x^2+1)(x-1)^2=ax^2, \ a\in\mathbb{R}$. Discuss.

This is the entire text of the problem. I am really unsure what the authors meant by discuss here? Is it to find $x$ based on $a$ and then separate by cases? What I tried doing is to raise to power and multiply the parentheses, but I'm not sure what to do next:
$$(x^2+1)(x-1)^2-ax^2 =0 \iff (x^2+1)(x^2-2x+1)-ax^2 = 0 \iff \\ \iff x^4 -2x^3+(2-a)x^2-2x+1 = 0$$
What I noticed is that this looks like a symmetric polynomial, would that help? And, again, how would I "discuss" that?
 A: $(x^2+1)(x-1)^2=ax^2\quad\color{blue}{(*)}$
Since $\,x=0\,$ is not a solution of the equation $(*)\,,\,$ we can divide by $\,x^2$ both sides of $(*)$ and get the following equivalent equation:
$\left(x+\dfrac1x\right)\left(x+\dfrac1x-2\right)=a\;.$
By letting $\;t=x+\dfrac1x\,,\,$ we get the following equation:
$t\,(t-2)=a\quad\color{blue}{(**)}$
which does not have any real solution if $\,a<-1\,.$
Whereas, if $\,a\geqslant-1\,,\,$ $(**)$ has the following real solutions :
$t=1\pm\sqrt{a+1}$.
Moreover, since $\;t=x+\dfrac1x\,,\,$ it follows that
$x+\dfrac1x=1\pm\sqrt{a+1}\;\;,$
$x^2-\left(1\pm\sqrt{a+1}\right)x+1=0\quad\color{blue}{(*\!*\!*)}$
which is equivalent to the equation $(*)$.
If $\;a<0\,,\,$ the equation $(*\!*\!*)$ does not have any real solution.
If $\;0\leqslant a<8\,,\,$ the equation $(*\!*\!*)$ has the following two real solutions:
$x=\dfrac{1+\sqrt{a+1}\pm\sqrt{a-2+2\sqrt{a+1}}}2\,.$
Whereas, if $\,a\geqslant8\,,\,$ the equation $(*\!*\!*)$ also has other two real solutions:
$x=\dfrac{1-\sqrt{a+1}\pm\sqrt{a-2-2\sqrt{a+1}}}2\,.$
A: This is not a full solution. Just pointers.
Start with $a\lt 0:\text{ no solution}$ since left side of equal is positive and the right side must be positive as well
Since you’re at it, try $a=0\Rightarrow x=1:\text{two real and two complex roots}$
Then move to $a\gt 0, x=\tan t:(x^2+1)(x-1)^2=ax^2\Leftrightarrow a\sin^2(2t)+4\sin(2t)-4=0$ and analyze $ay^2+4y-4=0,a\gt 0, -1\le y\le 1$
