At what points $c$ does the conclusion of the Mean Value Theorem hold for $(x)=x^{3}$ on the interval $(-4.4)$ ?

The conclusion of the Mean Value Theorem holds for $c=$
(Use a comma to separate answers as needed. Type an exact answer using radicals as needed)

Step 1 ：We are given the function \(f(x) = x^{3}\) and the interval (-4, 4).

Step 2 ：The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at that point is equal to the average rate of change of the function over the interval.

Step 3 ：The derivative of the function is \(f'(x) = 3x^{2}\).

Step 4 ：The average rate of change of the function over the interval is given by \(\frac{f(b) - f(a)}{b - a} = \frac{4^{3} - (-4)^{3}}{4 - (-4)} = 0\).

Step 5 ：We need to solve the equation \(f'(c) = 0\) for c in the interval (-4, 4).

Step 6 ：Solving the equation gives us the value of c as 0.

Step 7 ：Final Answer: The conclusion of the Mean Value Theorem holds for \(c=\) \(\boxed{0}\).