$1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\,\,(x\ne-2)\\⇔1+\dfrac{1}{x+2}-\dfrac{12}{(x+2)(x^2-2x+4)}=0\\⇔\dfrac{(x+2)(x^2-2x+4)+x^2-2x+4-12}{(x+2)(x^2-2x+4)}=0\\⇔x^3+8+x^2-2x-8=0\\⇔x^3+x^2-2x=0\\⇔x(x^2+x-2)=0\\⇔x(x-1)(x+2)=0\\⇔\left[\begin{array}{}\hspace{-0,8cm}x=0\\x-1=0\\x+2=0\end{array}\right.\\⇔\left[\begin{array}{}x=0\text{ (thoả mãn)}\\x=1\text{ (thoả mãn)}\\\hspace{-0,8cm}x=-2\text{ (loại)}\end{array}\right.$
