\(x^3-7x+6=0\)
\(\Leftrightarrow x^3-x-6x+6=0\)
\(\Leftrightarrow(x^3-x)-(6x-6)=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)-6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2+x-6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-3x+2x-6\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-2;1;3\right\}\)