\(\left(x+1\right)^3+\left(x-1\right)^3=\left(x-1\right)\left(x+1\right)+4\)

\(\Rightarrow\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]-\left(x+1\right)\left(x-1\right)-4=0\)

\(\Rightarrow2x\left(x^2+2x+1-x^2+1+x^2-2x+1\right)-x^2+1-4=0\)

\(\Rightarrow2x\left(x^2+3\right)-x^2+1-4=0\)

\(\Rightarrow2x^3+6x-x^2-3=0\)

\(\Rightarrow\left(2x^3+6x\right)-\left(x^2+3\right)=0\)

\(\Rightarrow2x\left(x^2+3\right)-\left(x^2+3\right)=0\)

\(\Rightarrow\left(x^2+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=-3\left(L\right)\\x=\dfrac{1}{2}\end{matrix}\right.\)