Ta có: \(VT=\left(x^4+x^3\right)-\left(3x^3+3x^2\right)+\left(x^2+x\right)+2\left(x+1\right)\)
\(=x^3\left(x+1\right)-3x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-2x^2-x^2+2x-x+2\right)\)
\(=\left(x+1\right)\left[x^2\left(x-2\right)-x\left(x-2\right)-\left(x-2\right)\right]\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2-x-1\right)\)
Do vậy pt tương đương với \(\left(x+1\right)\left(x-2\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\) . Giải cái ngoặc cuối cùng: \(x^2-x-1=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{5}}{2}\\x=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\)
