Here is the revised proof:
Proof:
Let f(x) = ax^3 + bx + c.
Assume that f(x) has 3 distinct roots, f(p) = f(q) = f(r) = 0, there is a point x1 element of (p,q) such that f'(x1) = 0 ; x2 element of (q, r) such that f'(x2) = 0.
Since ab < 0 then there are two possibilities where a>0 and b<0...