I have the following equation
$$xy^2-\frac{x(-xy-1)^2}{(x-1)^2}+2y\left(\frac{-xy-1}{x-1}-1\right)+\left(\frac{-xy-1}{x-1}-1\right)^2=0$$
which I was trying to simplify. I know the solution is $$\frac{x(y+1)(3y-1)}{x-1}=0$$ which gives $y=-1$ or $y=1/3$.
WolframAlpha suggests first expanding and collecting in terms of $y$:
$$1+\frac{1}{(x-1)^2}+\frac{2}{x-1}-\frac{x}{(x-1)^2}+y\left(-2-\frac{2}{x-1}+\frac{2x}{(x-1)^2}+\frac{2x}{x-1}-\frac{2x^2}{(x-1)^2}\right)+y^2\left(x-\frac{2x}{x-1}+\frac{x^2}{(x-1)^2}-\frac{x^3}{(x-1)^2}\right)=0$$
and then "factoring the left side into a product with five terms" i.e the solution. Is this factorization obvious? How do I factor the LHS by hand, or is there another simpler way?
Note that $\frac{-xy-1}{x-1}-1=-\frac{x(y+1)}{x-1}.$
So taking out a common factor x and putting every term over the common denominator $(x-1)^{2}$ we have:
$$x(\frac{y^{2}(x-1)^{2}-(-xy-1)^{2}-2y(y+1)(x-1)+x(y+1)^{2}}{(x-1)^{2}})=x(\frac{-3y^{2}(x-1)-2y(x-1)+(x-1)}{x-1})=\frac{x(-3y^{2}-2y+1)}{x-1}=-\frac{x(y+1)(3y-1)}{x-1}=0.$$