设f(x)=x³-x²-x,有f'(x)=3x²-2x-1=(3x+1)(x-1).

　　当x≥1或x≤-1/3时f'(x)≥0,故f(x)在(-∞,-1/3]和[1,+∞)上单调递增.

　　而当-1/3≤x≤1时,f'(x)≤0,故f(x)在[-1/3,1]上单调递减.

　　f(2)=2,故x≥2时f(x)≥2,

　　然而方程的根满足x+x²≥[x]+[x²]=[x³]>x³-1,即f(x)