1. Derive f(x) ` = 3x 2-a, if r increases, then f(x )` is constant >; =0, do a parabola, always above the X axis. Get a < =0.

2. If the range of (-1, 1) is reduced, the derivative f(x) ` must be less than 0 in this range. As can be seen from the figure, the derivative image should have an upward opening, and the abscissa intersecting with the X axis should satisfy X.

Then the root of 3x 2-a = 0 should satisfy the above X range, a >: = 3.

3. train of thought: prove that no matter what the value of a is, x 3-ax-1

x^3-ax & lt； a+ 1

Let y 1 = x 3-ax and y2 = a+ 1.

y 1` = 3x^2 -a，y2` =0

Y 1 is a odd function, which is symmetrical about the origin. The slope increases with X, and the slope does not approach 0, which increases infinitely. The range extends up and down indefinitely, and the range is all real numbers.

So no matter what value A takes, X can always be found to satisfy the inequality.