$1+\frac{1}{x+2}=$ $\frac{12}{8+x^3}$
<=>$\frac{x+2+1}{x+2}=$ $\frac{12}{8+x^3}$
<=>$\frac{x+3}{x+2}-$ $\frac{12}{x^3+2^3}=0$
<=>$\frac{(x+3)(x^2-2x+4)}{(x+2)(x^2-2x+4)}-$ $\frac{12}{(x+2)(x^2-2x+4)}=0$
<=>$(x+3)(x^2-2x+4)-12=0$
<=>$x^3-2x^2+4x+3x^2-6x+12-12=0$
<=>$x^3+x^2-2x=0$
<=>$x(x^2+x-2)=0$
<=>$x(x+2)(x-1)=0$
<=>$x=0;x+2=0;x-1=0$
<=>$x=0;x=-2;x=1$
Vậy S={0;-2;1}
