If the function $$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$$, $$a > 0$$ has a local maximum at $$x = \alpha$$ and a local minimum at $$x = \alpha^2$$, then $$\alpha$$ and $$\alpha^2$$ are the roots of the equation :

Given $$f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$$ with $$a > 0$$. Differentiating: $$f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a)$$. The critical points are $$x = a$$ and $$x = 2a$$.

To classify these, we use the second derivative $$f''(x) = 12x - 18a$$. At $$x = a$$: $$f''(a) = 12a - 18a = -6a$$. Since $$a > 0$$, we have $$f''(a) < 0$$, confirming a local maximum at $$x = a$$. At $$x = 2a$$: $$f''(2a) = 24a - 18a = 6a > 0$$, confirming a local minimum at $$x = 2a$$.

We are told the local maximum occurs at $$x = \alpha$$ and the local minimum at $$x = \alpha^2$$. So $$\alpha = a$$ and $$\alpha^2 = 2a$$. Substituting $$a = \alpha$$ into the second equation: $$\alpha^2 = 2\alpha$$, giving $$\alpha(\alpha - 2) = 0$$. Since $$\alpha = a > 0$$, we get $$\alpha = 2$$, and therefore $$\alpha^2 = 4$$.

The required equation has roots $$2$$ and $$4$$. Their sum is $$6$$ and product is $$8$$, so the equation is $$x^2 - 6x + 8 = 0$$, which is $$\boxed{\text{Option (A)}}$$.