Đặt x+1=a; x-2=b
Phương trình trở thành:
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)=\left(a+b\right)^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow x\in\left\{-1;2;\dfrac{1}{2}\right\}\)
\(\left(x+1\right)^3+\left(x-2\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-6x^2+12x-8=8x^3-12x^2+6x-1\)
\(\Leftrightarrow2x^3-3x^2+15x-7-8x^3+12x^2-6x+1=0\)
\(\Leftrightarrow-6x^3+9x^2+9x-6=0\)
\(\Leftrightarrow-3\left(2x^3-3x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(2x^3+2\right)-\left(3x^2+3x\right)=0\)
\(\Leftrightarrow2\left(x^3+1\right)-3x\left(x+1\right)=0\)
\(\Leftrightarrow2\left(x^2-x+1\right)\left(x+1\right)-3x\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+2-3x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-4x-x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(2x^2-4x\right)-\left(x-2\right)\right]\left(x+1\right)=0\)
\(\Leftrightarrow\left[2x\left(x-2\right)-\left(x-2\right)\right]\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(\text{Vậy tập nghiệm phương trình là:}\left\{\dfrac{1}{2};2;\left(-1\right)\right\}\)