$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.
 A: Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have
$$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3}  =1-\frac{3}{xy} \tag{1}$$
But we have also
$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$
From (1) and (2), we deduce
\begin{align}
&(\frac{1}{x}+\frac{1}{y})^3 = (1-\frac{3}{xy}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \\
&\iff (\frac{1}{x}+\frac{1}{y})^3- 1=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1   \right)=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1  -\frac{3}{xy} \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} +(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy}   \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}-\frac{1}{y})^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( ((1+\frac{1}{x})-(1+\frac{1}{y}))^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( \left(1+\frac{1}{x} \right)^2+\left(1+\frac{1}{y} \right)^2  -\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
\end{align}
The last equation happens if and only if $\frac{1}{x}+\frac{1}{y} =1$  or $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $. Hence, the two possible values of $\frac{1}{x}+\frac{1}{y}$ is $1$ and $-2$.
A: Hint:
Let $\dfrac{x+y}{xy}=k$
$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$
$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$
$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$
As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$
Clearly, $k=1$ is a solution.
Otherwise,  $$xy=\dfrac3{1+k+k^2}$$
A: $$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$
$$\frac{x+y}{xy}\tag{2}$$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$
can be factorized$^1$ as
$$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$
$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are only those of $x+y-xy=0$ that is $x+y=xy$ and finally
$$\frac{x+y}{xy}=1$$

$(^1)$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=(x+y)^3-3x^2y-3xy^2+ 3 x^2 y^2 - x^3 y^3=$$
$$=\left[(x+y)^3- x^3 y^3\right]+\left[-3x^2y-3xy^2+ 3 x^2 y^2 \right]=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)\right]-3 x y (x+y-x y)=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)-3xy\right]=(x+y-xy)\left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)$$
$(^2)$
$$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$$
set $y=kx$. We get
$$k^2 x^4+k^2 x^3+k^2 x^2+k x^3-k x^2+x^2=0\to x^2 \left(k^2 x^2+k^2 x+k x+k^2-k+1\right)=0$$
As $x\ne 0$ we have
$$k^2 x^2+k^2 x+k x+k^2-k+1=0$$
the discriminant of this equation is $D=-k^4+2 k^3-k^2=-k^2(k-1)^2$
$D\ge 0\to k=1 $ (solution $k=0$ is discarded)
if $k=1$ we have $x^2+2x+1=0\to x=-1$ thus the solutions are
$$x=-1;\;y=-1$$
A: The following is equivalent to @NN2's solution, only written a bit differently.
First substitute $x=1/a, y=1/b$, so that the equation becomes
$$ \tag{*}
 a^3+b^3+3ab - 1 = 0 \, .
$$
We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then
$$
 0 = a^3+b^3+3ab - 1 = S^3+3PS + 3P -1 = (S-1)(S^2 + S-3P+1) \, .
$$
So $S=1$ is one possible value for $a+b$. Conversely, all $(a, b)$ with $a+b=1$ satisfy the equation $(*)$.
In order to determine the possible zeros of the second factor we use that $P \le S^2/4$ from the inequality between geometric and arithmethic mean:
$$
0 = S^2 + S-3P+1 \ge S^2 + S-\frac 34 S^2+1 = \frac 14 (S+2)^2 \ge 0
$$
is only possible if $S=-2$ and $P=1$, i.e. if $(a, b) = (-1, -1)$.
Therefore $S=a+b=1/x+1/y$ can only take the values $1$ and $-2$, and both values do actually occur.
