If $$\frac{1}{\sqrt{\alpha}}$$, $$\frac{1}{\sqrt{\beta}}$$ are the roots of the equation $$ax^2 + bx + 1 = 0$$, $$(a \neq 0, a, b \in R)$$, then the equation $$x(x + b^3) + (a^3 - 3abx) = 0$$ has roots:

Given that $$\frac{1}{\sqrt{\alpha}}$$ and $$\frac{1}{\sqrt{\beta}}$$ are the roots of the equation $$ax^2 + bx + 1 = 0$$, where $$a \neq 0$$ and $$a, b$$ are real numbers. For a quadratic equation $$px^2 + qx + r = 0$$, the sum of roots is $$-\frac{q}{p}$$ and the product is $$\frac{r}{p}$$. Applying this to the given equation:

Sum of roots: $$\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = -\frac{b}{a}$$

Product of roots: $$\frac{1}{\sqrt{\alpha}} \cdot \frac{1}{\sqrt{\beta}} = \frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a}$$

From the product, $$\frac{1}{\sqrt{\alpha\beta}} = \frac{1}{a}$$, so $$\sqrt{\alpha\beta} = a$$. Squaring both sides, $$\alpha\beta = a^2$$. ...(i)

From the sum, $$\frac{1}{\sqrt{\alpha}} + \frac{1}{\sqrt{\beta}} = \frac{\sqrt{\beta} + \sqrt{\alpha}}{\sqrt{\alpha\beta}}$$. Substituting $$\sqrt{\alpha\beta} = a$$, we get $$\frac{\sqrt{\alpha} + \sqrt{\beta}}{a} = -\frac{b}{a}$$. Multiplying both sides by $$a$$, $$\sqrt{\alpha} + \sqrt{\beta} = -b$$. ...(ii)

Since $$\alpha$$ and $$\beta$$ are positive (as roots involve square roots), $$\sqrt{\alpha} > 0$$ and $$\sqrt{\beta} > 0$$, so their sum is positive. Thus, $$-b > 0$$, implying $$b < 0$$. Set $$p = \sqrt{\alpha}$$ and $$q = \sqrt{\beta}$$, so $$p > 0$$, $$q > 0$$. Then:

From (i): $$p^2 q^2 = a^2$$, so $$(p q)^2 = a^2$$. Since $$a > 0$$ (because the product of roots $$\frac{1}{a} > 0$$), $$p q = a$$. ...(iii)

From (ii): $$p + q = -b$$. ...(iv)

Now, consider the equation $$x(x + b^3) + (a^3 - 3abx) = 0$$. Expanding:

$$x \cdot x + x \cdot b^3 + a^3 - 3abx = x^2 + b^3 x + a^3 - 3abx = x^2 + (b^3 - 3ab)x + a^3 = 0$$. ...(v)

This is a quadratic in $$x$$. Let its roots be $$m$$ and $$n$$. Sum of roots $$m + n = -\text{(coefficient of } x\text{)} = -(b^3 - 3ab) = -b^3 + 3ab$$. Product $$m n = \text{constant term} = a^3$$.

Substitute $$a = p q$$ and $$b = -(p + q)$$ from (iii) and (iv):

Sum: $$3ab - b^3 = 3(p q) \cdot [-(p + q)] - [-(p + q)]^3 = -3 p q (p + q) - [- (p + q)^3] = -3 p q (p + q) + (p + q)^3$$.

Factor out $$(p + q)$$:

$$(p + q) [ -3 p q + (p + q)^2 ]$$.

Expand $$(p + q)^2 = p^2 + 2 p q + q^2$$:

$$(p + q) [ -3 p q + p^2 + 2 p q + q^2 ] = (p + q) (p^2 - p q + q^2)$$.

Now, $$p^2 - p q + q^2 = p^2 + q^2 - p q$$. But $$p^2 + q^2 = (p + q)^2 - 2 p q$$, so:

$$p^2 + q^2 - p q = (p + q)^2 - 2 p q - p q = (p + q)^2 - 3 p q$$.

Thus, sum $$= (p + q) [ (p + q)^2 - 3 p q ]$$.

Product $$= a^3 = (p q)^3$$.

Note that $$p^3 + q^3 = (p + q)(p^2 - p q + q^2) = (p + q) [ (p + q)^2 - 3 p q ]$$, which matches the sum. Also, $$p^3 q^3 = (p q)^3 = a^3$$, which matches the product. Therefore, the roots are $$p^3$$ and $$q^3$$.

Since $$p = \sqrt{\alpha}$$, $$p^3 = (\sqrt{\alpha})^3 = \alpha^{\frac{3}{2}}$$. Similarly, $$q^3 = \beta^{\frac{3}{2}}$$.

Now, check the options:

A. $$\sqrt{\alpha\beta}$$ and $$\alpha\beta$$: These are $$a$$ and $$a^2$$, but roots are $$\alpha^{\frac{3}{2}}$$ and $$\beta^{\frac{3}{2}}$$, not matching.

B. $$\alpha^{-\frac{3}{2}}$$ and $$\beta^{-\frac{3}{2}}$$: These are reciprocals, not matching.

C. $$\alpha\beta^{\frac{1}{2}}$$ and $$\alpha^{\frac{1}{2}}\beta$$: These are $$\alpha \sqrt{\beta}$$ and $$\sqrt{\alpha} \beta$$, which are $$p^2 q$$ and $$p q^2$$. Sum is $$p^2 q + p q^2 = p q (p + q) = a (-b)$$, but required sum is $$p^3 + q^3$$, not equal in general.

D. $$\alpha^{\frac{3}{2}}$$ and $$\beta^{\frac{3}{2}}$$: These are $$p^3$$ and $$q^3$$, matching the roots.

Hence, the correct answer is Option D.