Ta có: \(x^3-12=13x\)
\(\Leftrightarrow x^3-13x-12=0\)
\(\Leftrightarrow x^3-x-12x-12=0\)
\(\Leftrightarrow x\left(x^2-1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)-12\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-12\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+3x-12\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\\x=-3\end{matrix}\right.\)
Vậy: S={-1;4;-3}
