Ta có: \(x^3-7x-6=0\)
\(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-1\right)-6\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;-2;3\right\}\)
