If the set of all values of a, for which the equation $$5x^{3}-15x-a=0$$ has three distinct real roots, is the interval $$(\alpha, \beta)$$, then $$\beta-2\alpha $$ is equal to ______

We need to find the interval $$(\alpha, \beta)$$ of values of $$a$$ for which $$5x^3 - 15x - a = 0$$ has three distinct real roots, then compute $$\beta - 2\alpha$$.

$$5x^3 - 15x = a$$, i.e., we need the horizontal line $$y = a$$ to intersect the curve $$y = 5x^3 - 15x$$ at three distinct points.

Differentiate: $$f'(x) = 15x^2 - 15 = 15(x^2 - 1)$$.

Setting $$f'(x) = 0$$: $$x^2 - 1 = 0$$, so $$x = \pm 1$$.

$$f''(x) = 30x$$.

At $$x = -1$$: $$f''(-1) = -30 < 0$$, so $$x = -1$$ is a local maximum.

At $$x = 1$$: $$f''(1) = 30 > 0$$, so $$x = 1$$ is a local minimum.

Local maximum: $$f(-1) = 5(-1)^3 - 15(-1) = -5 + 15 = 10$$.

Local minimum: $$f(1) = 5(1)^3 - 15(1) = 5 - 15 = -10$$.

The cubic $$y = f(x)$$ has a local max of 10 and a local min of $$-10$$. For the line $$y = a$$ to cut the curve at exactly three distinct points, $$a$$ must lie strictly between the local minimum and local maximum values:

$$-10 < a < 10$$

Therefore $$\alpha = -10$$ and $$\beta = 10$$.

$$\beta - 2\alpha = 10 - 2(-10) = 10 + 20 = 30$$

The correct answer is 30.