\(x^3+1=x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)-x\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)
Mà \(x^2+1>0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
Vậy x = 1