Does the value belong to a certain range or just finite numbers? Cause my answer belongs to a range PS: Could you please reply my first post? It's about combinatorics

FrauEuler wrote: Does the value belong to a certain range or just finite numbers? Cause my answer belongs to a range PS: Could you please reply my first post? It's about combinatorics no it's a pure algebra problem if i remember well the answers are $-2$ and $1$

Some progress: Note that $(x+y)^3 = x^3y^3 - 3x^2y^2 + 3xy^2 + 3x^2y$. Dividing through by $xy$ and letting $\frac{1}{x} + \frac{1}{y} = a$, we see that $$(x+y)^2a = x^2y^2 - 3xy + x + y.$$Dividing through by $xy$ again and rearranging $(x+y)a^2 - a = xy - 3$. Noting that $xy = \frac{x+y}{a}$ and multiplying both sides by $a$, we see that $(x+y)a^3 - a^2 = x+y - 3a$. Letting $x+y = b$ and rearranging, we find that $$ba^3 - a^2 + 3a - b = 0.$$

solution: notice that $(x+y)^3-x^3y^3=x^3+y^3+3x^2y+3xy^2-x^3y^3=-x^2y^2+3x^2y+3xy^2=3xy(-xy+x+y)$ or $(x+y)^3-x^3y^3=(x+y-xy)((x+y)^2+xy(x+y)+x^2y^2)=(x+y-xy)(x^2+y^2+2xy+x^2y+xy^2+x^2y^2)$ hence $x+y-xy=0$ or $x^2+y^2+x^2y+xy^2+x^2y^2+2xy=3xy$ the second is ie to $(x-y)^2+(x(y+1))^2+(y(x+1))^2=0$ implies $x=y=-1$ the first equality implies that $\frac{1}{x}+\frac{1}{y}=1$ hence the values are $1$ and $-2$

It's just $a^3+b^3+c^3-3abc=0$ or $(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0$ applied for the numbers $(x,y,-xy)$. Easy to find after this the only values: $-2,1$

GGPiku wrote: It's just $a^3+b^3+c^3-3abc=0$ or $(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0$ applied for the numbers $(x,y,-xy)$. Easy to find after this the only values: $-2,1$ Wow I really don't know that there can be such an easy solution with that formula

We have \begin{align*} &x^3+y^3+3x^2y^2=x^3y^3 \\ \iff &\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}-1=0 \\ \iff& \left(\frac{1}{x}+\frac{1}{y}\right)^3-1-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}-1\right)=0 \\ \iff & \left(\frac{1}{x}+\frac{1}{y}-1\right)\left(\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}+\frac{1}{x}+\frac{1}{y}+1\right)=0 \\ \iff & \frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}-1\right)\left(\left(\frac{1}{x}-\frac{1}{y}\right)^2+\left(\frac{1}{x}+1\right)^2+\left(\frac{1}{y}+1\right)^2\right)=0. \end{align*}Therefore, $\frac{1}{x}+\frac{1}{y}=1$ or $-2$. $\blacksquare$

Medjl wrote: let $x,y$ be non-zero reals such that : $x^3+y^3+3x^2y^2=x^3y^3$ find all values of $\frac{1}{x}+\frac{1}{y}$ This is straightforward if you love $a^3 + b^3 + c^3 = 3abc$. Note that \[ x^3 + y^3 + (-xy)^3 = 3(x)(y)(-xy) \]This forces $x + y - xy = 0$ or $x = y = -xy$. The former case gives us $\frac{1}{x} + \frac{1}{y} = 1$. The latter case gives us $x = y = -1$, and hence $\frac{1}{x} + \frac{1}{y} = -2$.