F(x) is continuous on [0, 1], f (0) = 1 > 0, f (1) =- 1 < 0. From the zero point theorem, there exists ξ∈ (0, 0.

Within (0, 1), f' (x) = 3x 2-3 < 0, and f(x) decreases monotonically, so if there is a root, it must be unique.

Therefore, f(x) has only one real root on [0, 1].