How to solve an equation of the form $f(x)=f(a)$ for a fixed real a. I got stuck on this question: find all solutions $x$ for $a\in R$:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{(a^2-a+1)^3}{a^2(a-1)^2}$$
I see that if we simplify we get:
$$\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{[(x-{\frac 12})^2+{\frac 34}]^3}{[(x-{\frac 12})^2-{\frac 14}]^2}$$
From the expression $(x-{\frac 12})^2$, I see that if $x=x_1$ is a solution, then $x=1-x_1$ is also a solution. But in the solution to this exercise, it was stated that $x=\frac{1}{x_1}$ must also be a solution, and I don't see how.
[EDIT]
Ok, thx for the help guys. What do you think of this solution (doesn't involve any above precalculus math, and needs no long calculations)?
From the above we know that if $x_1=a$ is a solution, then $x_2=1-a$ is also a solution.
Also, from here:
$$\require{cancel}\frac{(x^2-x+1)^3}{x^2(x-1)^2}=\frac{\cancel{x^3}(x+{\frac 1x}-1)^3}{\cancel{x^3}(x+{\frac 1x}-2)}$$
in the expression $x+{\frac 1x}$ we see that if $x=x_1$ is a solution, then $x=\frac{1}{x_1}$ is also a solution, so $x_3=\frac{1}{a}$.
With these two rules we can now keep generating roots until we have 6 total.
If $x=x_2$ is a solution, then $x=\frac{1}{x_2}$ is also a solution, so $x_4=\frac{1}{1-a}$.
If $x=x_3$ is a solution, then $x=1-x_3$ is also a solution, so $x_5=\frac{a-1}{a}$.
Finally, if $x=x_5$ is a solution, then $x=\frac{1}{x_5}$ is also a solution, so $x_6=\frac{a}{a-1}$
The 6 obtained values are distinct, so they cover all the roots.
[EDIT2]
I guess this is answered. No sure whose particular answer to actually select as the right one since they're all correct, so I'll just leave it like this.

factorizing the given equation and cancel the denominators we get
$$(ax-a-x)(ax-x+1)(x-1+a)(-x+a)(ax-1)(ax-a+1)=0$$
