\(\left(x+1\right)^2\left(x+2\right)+\left(x-1\right)^2\left(x-2\right)=12\)

\(\Leftrightarrow\left(x^2+2x+1\right)\left(x+2\right)+\left(x^2-2x+1\right)\left(x-2\right)=12\)\(\Leftrightarrow x^3+2x^2+2x^2+4x+x+2+x^3-2x^2-2x^2+4x+x-2-12=0\)\(\Leftrightarrow2x^3+10x-12=0\)

\(\Leftrightarrow2x^3-2x+12x-12=0\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+6\right)=0\)

Ta có :

\(x^2+x+6=x^2+x+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\)\(\Rightarrow x-1=0\Leftrightarrow x=1\)

Vậy phương trình có tập nghiệm \(S=\left\{1\right\}\)