\(\left\{{}\begin{matrix}x^3-8x=y^3+2y\left(1\right)\\x^2-3=3\left(y^2+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=8x+2y\\x^2-3y^2=6\end{matrix}\right.\)
\(\Rightarrow6\left(x^3-y^3\right)=\left(x^2-3y^2\right)\left(8x+2y\right)\)
\(\Leftrightarrow6x^3-6y^3=8x^3-24xy^2+2x^2y-6y^3\)
\(\Leftrightarrow2x^3-24xy^2+2x^2y=0\)
\(\Leftrightarrow2x\left(x-3y\right)\left(x+4y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3y\\x=-4y\end{matrix}\right.\)
TH1: \(x=0\)
\(\left(1\right)\Leftrightarrow y^3+2y=0\)
\(\Leftrightarrow y=0\)
TH2: \(x=3y\)
\(\left(1\right)\Leftrightarrow27y^3-24y=y^3+2y\)
\(\Leftrightarrow26y^3-26y=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=-1\end{matrix}\right.\)
Với \(y=0\Rightarrow x=0\)
Với \(y=1\Rightarrow x=3\)
Với \(x=-1\Rightarrow y=-3\)
TH3: \(x=-4y\)
\(\left(1\right)\Leftrightarrow-64y^3+32y=y^3+2y\)
\(\Leftrightarrow65y^3-30y=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=\pm\dfrac{\sqrt{78}}{13}\end{matrix}\right.\)
Với \(x=0\Rightarrow y=0\)
Với \(x=\dfrac{\sqrt{78}}{13}\Rightarrow y=-\dfrac{4\sqrt{78}}{13}\)
Với \(x=-\dfrac{\sqrt{78}}{13}\Rightarrow y=\dfrac{4\sqrt{78}}{13}\)
Đến đây thử lại rồi kết luận
