Đặt: \(x+2=t\)
\(pt\Leftrightarrow\left(t+1\right)^3-\left(t-1\right)^3=56\)
\(\Rightarrow\left(t^3+3t^2+3t+1\right)-\left(t^3-3t^23t-1\right)=56\)
\(\Rightarrow t^3+3t^2+3t+1-t^3+3t^2-3t+1=56\)
\(\Rightarrow\left(t^3-t^3\right)+\left(3t^2+3t^2\right)+\left(3t-3t\right)+\left(1+1\right)=56\)
\(\Rightarrow6t^2+2=56\Leftrightarrow6t^2=54\Leftrightarrow t^2=9\)
\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)