Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 24

You need to notice that the given function is continuous on [0,1] and differentiable on (0,1), since it is a polynomial function.
You need to verify if f(0)=f(1), hence, you need to evaluate the values of function at x = 0 and x = 1.
f(0) = 0 - 0^(1/3) = 0
f(1) = 1 - 1^(1/3) = 1-1 = 0
Since f(0)=f(1) = 0 and the function is continuous and differentiable on the given interval, the Rolle's theorem may be applied, hence, there is a point c in (0,1), such that:
f'(c)(1 - 0) = 0
You need to find the derivative of the function:
f'(c) = (c - c^(1/3))' => f'(c) = 1 - (1/3)c^(1/3-1)
f'(c) = 1 - 1/(2*root(3)(c^2))
Replacing the found values in equation f'(c)(1 - 0) = 0 yields:
1 - 1/(2*root(3)(c^2)) = 0 => 1/(2*root(3)(c^2)) = 1 => (2*root(3)(c^2)) = 1 => root(3)(c^2) = 1/2
Raise to 3rd power both sides:
c^2 = (1/2)^3 => c^2 = 1/8 => c_1 = sqrt 1/(2sqrt2); c_2 = -1/(2sqrt2)
Notice that c_2 = -1/(2sqrt2) does not belong to (0,1).
Hence, applying Rolle's theorem to the given function yields that there is c = sqrt 1/(2sqrt2) in (0,1) , such that f'(c) = 0.