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