$x^3+y^3+3x^2y^2-x^3y^3=(x+y-xy) \frac{(x-y)^2+(x+xy)^2+(y+xy)^2}{2}=0$ So $x+y=xy \to \frac{1}{x}+\frac{1}{y}=1$ or $x=y=-xy \to x=y=-1 \to \frac{1}{x}+\frac{1}{y}=-2$

$~~~~~~$

Attachments:

samrocksnature wrote: $~~~~~~$ what is even that lol, i used gauss factorization (oops, i forgot that in usa u guys call it titu's factorization)

MathLuis wrote: samrocksnature wrote: $~~~~~~$ what is even that lol, i used gauss factorization (oops, i forgot that in usa u guys call it titu's factorization) Any handouts about this factorization method?