Đáp án:
\[{\color{Red} {\left\{\begin{matrix} x^3-y^3-3x^2-y+2=0\qquad(1) & \\ 3x^2+y^2-6x-1=0 \qquad(2)& \end{matrix}\right.}}\]
\[(2)↔3(x-1)^2+y^2=4\qquad(3)\]
\[↔(x-1)^2=\dfrac{4-y^2}{3}\]
\[(1)↔x^3-3x^2+2=y^3+y\]
\[↔(x-1)\bigg[(x-1)^2-3\bigg]=y^3+y\]
\[↔x-1\Bigg(\dfrac{4-y^2}{3}-3\Bigg)=y^3+y\]
\[↔(x-1).\dfrac{-y^2-5}{3}=y^3+y\]
\[↔x-1=\dfrac{-3(y^3+y)}{y^2+5}\]
Thay vào $(3),$ ta được:
\[3.\dfrac{9(y^3+y)^2}{(y^2+5)^2}+y^2=4\]
\[↔(y^2-1)(28y^4+88y^2+100)=0\]
\[↔y^2=1\]
\[↔y=±1\]
$→x=0; x=2$
Thử lại $→(x;y)=(0,1);(2;-1)$