\(\Leftrightarrow\left(x-1\right)^3=y^3+2y^2+1\)
Ta có: \(y^3+2y^2+1>y^3\)
\(y^3+2y^2+1=\left(y+4\right)^3-10\left(x+\frac{12}{5}\right)^2-\frac{27}{5}< \left(y+4\right)^3\)
\(\Rightarrow y^3< y^3+2y^2+1< \left(y+4\right)^3\)
\(\Rightarrow\left[{}\begin{matrix}y^3+2y^2+1=\left(y+1\right)^3\\y^3+2y^2+1=\left(y+2\right)^3\\y^3+2y^2+1=\left(y+3\right)^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2+3y=0\\4y^2+12y+7=0\\7y^2+27y+26=0\end{matrix}\right.\) \(\Rightarrow y=\left\{-3;-2;0\right\}\)
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