$3x^3 – 48x = 0$
⇔ $3x(x^2-16)=0$
⇔ \(\left[ \begin{array}{l}x=0\\x^2-16=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x^2-4^2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\(x-4)(x+4)=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=4;x=-4\end{array} \right.\)
Vậy $S=${$-4;0;4$}