\(\left\{{}\begin{matrix}x+y^2=y^3\left(1\right)\\y+x^2=x^3\left(2\right)\end{matrix}\right.\)
\(\Leftrightarrow x+y^2-\left(y+x^2\right)=y^3-x^3\)
\(\Leftrightarrow x+y^2-x^2-y+x^3-y^3=0\)
\(\Leftrightarrow x-y-\left(x^2-y^2\right)+x^3-y^3=0\)
\(\Leftrightarrow x-y-\left(x-y\right)\left(x+y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-x+y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\left(3\right)\\x^2+xy+y^2-x+y+1=0\left(4\right)\end{matrix}\right.\)
từ (1)(3) ta có hệ \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x+y^2=y^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y^3-y^2-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\\left[{}\begin{matrix}y=0\\y=\dfrac{1+\sqrt{5}}{2}\\y=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\y=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\\y=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
tương tự với (1)(4)
