$\begin{cases}x^3=2x+y(1)\\y^3=2y+x(2)\end{cases}$
$(1)-(2)⇔x^3-y^3=2y-2x+x-y$
$⇔(x-y)(x^2+xy+y^2)=y-x$
$⇔(x-y)(x^2+xy+y^2+1)=0$
Ta có:
$x^2+xy+y^2+1$
$=x^2+2.x.\dfrac{y}{2}+\dfrac{y^2}{4}+\dfrac{3y^2}{4}+1$
$=\bigg{(}x+\dfrac{y}{2}\bigg{)}^2+\dfrac{3}{4}y^2+1>0$
$Pt ⇔x-y=0$
$⇔x=y$
Thay vào $(1)$
$⇔x^3=2x+x$
$⇔x(x^2-3)=0$
$⇔\left[ \begin{array}{l}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{array} \right.$
