Find all possible values of $\frac{1}{x}+\frac{1}{y}$, if $x,y$ are real numbers not equal to $0$ that satisfy $$x^3+y^3+3x^2y^2=x^3y^3$$
Problem
Source: Bolivia IMO TST 2022 Day 1 P1
Tags: Bolivia, algebra, TST, Gauss identity
RagvaloD
28.07.2022 02:54
$x^3+y^3+3x^2y^2-x^3y^3=(x+y-xy) \frac{(x-y)^2+(x+xy)^2+(y+xy)^2}{2}=0$ So $x+y=xy \to \frac{1}{x}+\frac{1}{y}=1$ or $x=y=-xy \to x=y=-1 \to \frac{1}{x}+\frac{1}{y}=-2$
MathLuis
28.07.2022 03:56
samrocksnature wrote: $~~~~~~$ what is even that lol, i used gauss factorization (oops, i forgot that in usa u guys call it titu's factorization)
Cusofay
30.06.2023 16:11
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?