a, $x^3+3x^2+3x+2=0$
$⇔x^3+2x^2+x^2+2x+x+2=0$
$⇔x^2(x+2)+x(x+2)+(x+2)=0$
$⇔(x+2)(x^2+x+1)=0$
$+,x+2=0$
$⇔x=-2$
$+,x^2+x+1=0$
$⇔(x+\dfrac{1}{2})+\dfrac{3}{4}=0$ (vô lý)
Vậy `S={-2}`
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b, $x^3-12x^2+48x-72=0$
$⇔x^3-6x^2-6x^2+36x+12x-72=0$
$⇔x^2(x-6)-6x(x-6)+12(x-6)=0$
$⇔(x^2-6x+12)(x-6)=0$
$+,x^2-6x+12=0$
$⇔(x-3)^2+3=0$ (vô lý)
$+,x-6=0$
$⇔x=6$
Vậy `S={6}`