I am trying to solve this system of two cubic equations
$$\begin{cases}2x-yx^2-y^3 = 0\\ 2y-xy^2-x^3 = 0 \end{cases}$$
Is there a good way of attacking this in order to be sure to find ALL possible solutions? It is trivial to see that (0,0) and (1,1) are solutions, but I am at a loss trying to find other solutions.
Is a correct conclusion that since $x$ and $y$ have the same "role" in both equations that all solutions should be on the $y=x$ line? Or is there more to it?
Best Answer
Subtract the second one from the first. You get $2(x-y)-xy(x-y)+x^{3}-y^{3}=0$. If $x \neq y$ you can divide by $x-y$ and simplify to get $2+x^{2}+y^{2}=0$. There are no real numbers satisfying this (becasue the left side is $>0$). Hence we must have $x=y$ and now it is very easy to find all solutions.
[Apart from you solutions you get just one more solution: $(-1,-1)$].