What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors
 A: HINT Divide by $x^3y^3$ on both sides. You will get a new equation in terms of $1/x$ and $1/y.$ Can you finish from here?
A: The equation is equivalent to
$$x^3+y^3+(-xy)^3 -3xy(-xy)=0 \tag{1}$$
Applying this formula
$$a^3+b^3+c^3 -3abc= \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
for $(a,b,c) = (x,y,-xy)$
then $(1)$ holds true if and only if either of these 2 equations holds true $$x+y-xy = 0\tag{2}$$ $$x=y=-xy \tag{3}$$
If $(2)$ holds true then $\frac{1}{x}+\frac{1}{y} = 1$
if $(3)$ holds true then $x = y = -1 \implies \frac{1}{x}+\frac{1}{y} = -2$
Conclusion:
$$\frac{1}{x}+\frac{1}{y} = 1 \text{ or } \frac{1}{x}+\frac{1}{y} =-2$$
A: A solution proceeds as follows: Denote $t:=(1/x)+(1/y)$. Observe that $$t^3=\left(\frac{1}{x}+\frac{1}{y}\right)^3=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\cdot t.$$ For the given equation, you may divide $x^3y^3$ on both sides, which gives $$1=\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}.$$ Subtracting these two formulas, we can obtain $$t^3-1=\frac{3(t-1)}{xy}.$$ If $t=1$, this formula holds definitely. If $t\neq 1$, then $$t^3-1=\frac{3(t-1)}{xy}\implies t^2+t+1=\frac{3}{xy}\implies\frac{1}{x}\cdot\frac{1}{y}=\frac{t^2+t+1}{3}.$$ If you regard $u:=1/x$ and $v:=1/y$, then $u$ and $v$ are solutions to the quadratic equation $$z^2-tz+\frac{t^2+t+1}{3}=0.$$ Since $x$ and $y$ are real numbers, both $u$ and $v$ are also real numbers. This implies that the discriminant of this quadratic equation is non-negative. Thus $$\Delta=t^2-\frac{4(t^2+t+1)}{3}=-\frac{(t+2)^2}{3}\geq 0, $$ hence $t=-2$ if $t\neq 1$.
In general, $t=-2$ or $t=1$.
