$(x-1).(x^2+5x-2)-x^3+1=0$
$⇔(x-1).(x^2+5x-2)-(x-1)(x^2+x+1)=0$
$⇔(x-1)(x^2+5x-2-x^2-x-1)=0$
$⇔(x-1)(4x-3)=0$
$⇔\left[ \begin{array}{l}x-1=0\\4x-3=0\end{array} \right.⇔\left[ \begin{array}{l}x=1\\x=\frac{3}{4}\end{array} \right.$
Vậy $S=${$1;\frac{3}{4}$}.