Simplify $x^2 (1-y)=y^2 (1-x)$ It seems that $(x^2)(1-y) = (y^2)(1-x)$ should simplify. Wolfram Alpha says that solutions are $y=x$ and $y=1/(1-x)$. The first seems intuitively true, but I can't see the algebraic path from the original equation to the simplified solutions. Can anyone show me step by step how to get from the original to y=x? 
This question evolved out of another (I was trying to find a simpler way out). The original original question was $(x^{-2} + y^{-3}) / (x^{-3} + y^{-2}) = 1$  which also seems like it should simplify to $y=x$, but I cannot find the algebraic path to get there. I get stuck at $(x^3) / (1-x) = (y^3) / (1-y)$.
 A: $$
x^2(1-y) = y^2(1-x) = x^2 - x^2y = y^2 -xy^2 \quad \Rightarrow \quad x^2 - y^2 = x^2y - xy^2 \quad \Rightarrow
$$
$$
(x - y)(x + y) = -xy(x - y). \ \text{If} \ x \neq y \quad \Rightarrow \quad x + y = -xy \ \Rightarrow \ x + xy + y = 0  
$$
A: $$x^2(1-y)=y^2(1-x)$$
$$y^2(1-x)+x^2y-x^2=0$$
$$y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
where
$$a=1-x$$
$$b=x^2$$
$$c=-x^2$$
A: Plugging $y = x$ into
$$x^2 (1-y) = y^2 (1-x)$$
gives
$$x^2(1-x) = x^2(1-x)$$
which is true, therefore $y = x$ is a solution to the equation for $y$.
Don't do things the hard way if you don't have to: if you have a guess for a solution, you can simply check the solution rather than trying to derive it.
Now, if you want to know the complete solution space, here is an approach.
Now, expanding the equation and pulling everything to one side gives
$$ (1-x)y^2 + x^2 y - x^2 = 0$$
Because we know $y=x$ is a solution, we know that we can factor $(y-x)$ it out of the left hand side, and thus get the other solution for $y$.
Or we can do something simpler: we know that the two roots of the polynomial $ay^2 + by + c$ add up to $-b/a$ and multiply to give $c/a$.
Thus, the other solution for $y$ satisfies
$$ x y = \frac{-x^2}{1-x} $$
(aside: if you use this approach, you should treat the case $x=1$ separately. The approach requires that the polynomial be quadratic in $y$, but that is only true when $x \neq 1$).
