x,y are real numbers different from 0 such that :$x^3+y^3+3x^2y^2=x^3y^3$ Find all possible values of E=$\dfrac{1}{x}+\dfrac{1}{y}$
Problem
Source: Third Romanian JBMO TST 2016
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den_thewhitelion
15.06.2016 10:30
Hint: $x^3+y^3+(-xy)^3-3xy(-xy)=0$ and this is a formula...
Reynan
15.06.2016 12:25
$x^3+y^3+3x^2y^2=x^3y^3\Leftrightarrow (x+y-xy)(x^2+y^2+x^2y^2-xy+x^2y+xy^2)=0$ if $x+y-xy=0\implies E=1$ now $x^2+y^2+x^2y^2-xy+x^2y+xy^2=0\Leftrightarrow (x-y)^2+(x+xy)^2+(y+xy)^2=0$ and $x,y\not =0$ yield $x=y=-1$ hence $E=-2$
Reynan
15.06.2016 12:25
den_thewhitelion wrote: Hint: $x^3+y^3+(-xy)^3-3xy(-xy)=0$ and this is a formula... dont give hint
trigadd123
29.01.2021 13:56
Let us denote $S=x+y$ and $P=xy$. We want to compute $E=\frac{S}{P}$. Observe that
$$x^3+y^2+3x^2y^2=x^3y^3\Leftrightarrow S^3-3PS+3P^2=P^3$$$$\Leftrightarrow S^3-P^3=3PS-3P^2\Leftrightarrow (S-P)(S^2+SP+P^2)=3P(S-P)$$therefore we either have $S=P$, which gives $E=1$, or we can write
$$S^2+SP+P^2=3P$$$$\Leftrightarrow (3S^2-12P)+(S^2+4PS+4P^2)=0$$$$\Leftrightarrow 4(S^2-4P)+(S+2P)^2=0$$$$\Leftrightarrow 4(x-y)^2+(x+y+2xy)^2=0$$which gives that $x=y$ and $x+y+2xy=0$, or, since $x, y\neq 0$, $x=y=-1$ and thus $E=-2$.
We finally conclude that $E\in\left\{-2, 1\right\}$.
TuZo
29.01.2021 16:25
Reynan wrote: den_thewhitelion wrote: Hint: $x^3+y^3+(-xy)^3-3xy(-xy)=0$ and this is a formula... dont give hint Why not? A good hint (like this) it is better than a solution!