\(1+\dfrac{1}{2+x}=\dfrac{12}{x^3+8}\)
\(\Leftrightarrow1+\dfrac{1}{2+x}-\dfrac{12}{x^3+8}=0\)
\(\Leftrightarrow1+\dfrac{1}{x+2}-\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)+\left(x^2-2x+4\right)-12=0\)
\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8+x^2-2x+4-12=0\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+x-1-1\right)=0\)
\(\Leftrightarrow x\left[\left(x^2-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left[\left(x-1\right)\left(x+1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;1;-2\right\}\)