\(y\left(y+1\right)\left(y^2+y+1\right)\Leftrightarrow\left[y\left(y+1\right)\right]\left[y\left(y+1\right)+1\right]=42\)
\(Đặt:y\left(y+1\right)=t.Taco:t\left(t+1\right)=42\Leftrightarrow t^2+t+\frac{1}{4}=\frac{169}{4}\Leftrightarrow\left(t+\frac{1}{2}\right)^2=\frac{169}{4}\Leftrightarrow\left[{}\begin{matrix}t+\frac{1}{2}=-\frac{13}{2}\\t+\frac{1}{2}=\frac{13}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-7\\t=6\end{matrix}\right.\)\(+,t=-7\Rightarrow y\left(y+1\right)=-7\Leftrightarrow y^2+y=-7\Leftrightarrow y^2+y+\frac{1}{4}=-\frac{27}{4}\Leftrightarrow\left(y+\frac{1}{2}\right)^2=-\frac{27}{4}\left(\text{ loại}\right)\)
\(+,t=6\Rightarrow y^2+y=6\Leftrightarrow y^2+y+\frac{1}{4}=\frac{25}{4}\Leftrightarrow\left(y+\frac{1}{2}\right)^2=\frac{25}{4}\Leftrightarrow\left[{}\begin{matrix}y+\frac{1}{2}=-\frac{5}{2}\\y+\frac{1}{2}=\frac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-3\\y=2\end{matrix}\right..Vậy:y\in\left\{-3;2\right\}\)
