How to find solutions to this equality $\; \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$ We have the following equality:
$$ \mathrm{x} = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
Some of the solutions I found: 


*

*$\mathrm{x} = 0$ 

*Also for $\mathrm{a}=0$, every $\mathrm{x}$ is a solution I believe
I tried getting everything out of the brackets but that just gave a nasty equality which I couldn't solve. 
Let's take $-1 \leq \mathrm{a} \leq 1$. I was also wondering if there is a way of knowing how many solutions this equality has beforehand? Or do we just have to look at the cases $\mathrm{a} = 1$, $\mathrm{a} = -1$, $\mathrm{a} \neq 0$ and $\mathrm{a} = 0$ individually to find every solution (i.e. both of the bounds, and for a (not) equal to 0)?

I've solve upto some extent
$$ x = \mathrm{a^2x \, (1-x)\,(1-ax\,(1-x))}$$
$$ {a^2x(1-x)(1-ax(1-x))-x=0}\implies x=0$$
$$ {a^2(1-x)(1-ax(1-x))-1=0}$$
$$ {a^2(x-1)(1+ax(x-1))+1=0}$$
put x-1 =t
$$ {a^2t(1+at(t+1))+1=0}$$
$$ {a^2t(1+at^2+at)+1=0}$$
$$a^3t^3+a^3t^2+a^2t+1=0$$
for this equation if we take a=1
$$t^3+t^2+t+1=0\implies t=-1,i,-i\implies x=0,1+i,1-i$$
so based on value of a there are different solution of t and x.
solution in continuity based on @maming's comment 
$$a^3t^3+a^3t^2+a^2t+1=0$$
$$a^3t^3+1+a^3t^2+a^2t=0$$
$${(at)}^3+1^3+a^3t^2+a^2t=0$$
$$(at+1)(a^2t^2-at+1)+a^2t(at+1)=0$$
$$(at+1)(a^2t^2-at+1+a^2t)\implies (at+1)=0\;,(a^2t^2-at+1+a^2t)=0$$
$$t=\frac{-1}{a}\implies \mathbf{x=1-\frac 1a}\;\;,a^2t^2+t(a^2-a)+1=0$$
$$t=\dfrac {-(a^2-a)\pm\sqrt{{(a^2-a)}^2-4\cdot a^2\cdot 1}}{2a^2}$$
$$t=\dfrac {a\left((1-a)\pm\sqrt{{(a-1)}^2-4}\right)}{2a^2}$$
$$t=\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}\implies x=1+\dfrac {(1-a)\pm\sqrt{{(a-1)}^2-4}}{2a}$$
$$x=\dfrac {(a+1)\pm\sqrt{{(a-1)}^2-4}}{2a}$$
