If $$\alpha$$ and $$\beta$$ be two roots of the equation $$x^2 - 64x + 256 = 0$$. Then the value of $$\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8}$$ is:

First, simplify the expression completely in terms of $$\alpha + \beta$$ and $$\alpha\beta$$, and only then substitute their values.

Given:

$$\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8}$$

Step 1: Simplify the first term

To introduce $$\alpha\beta$$, multiply the numerator and denominator inside the fraction by $$\alpha^5$$:

$$\frac{\alpha^3}{\beta^5} = \frac{\alpha^3 \cdot \alpha^5}{\beta^5 \cdot \alpha^5} = \frac{\alpha^8}{(\alpha\beta)^5}$$

Therefore:

$$\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} = \left(\frac{\alpha^8}{(\alpha\beta)^5}\right)^{1/8} = \frac{\alpha}{(\alpha\beta)^{5/8}}$$

Similarly:

$$\left(\frac{\beta^3}{\alpha^5}\right)^{1/8} = \frac{\beta}{(\alpha\beta)^{5/8}}$$

Step 2: Add the terms

The whole expression becomes:

$$\frac{\alpha}{(\alpha\beta)^{5/8}} + \frac{\beta}{(\alpha\beta)^{5/8}}$$

Combining over the common denominator:

$$= \frac{\alpha+\beta}{(\alpha\beta)^{5/8}}$$

So the simplified form in terms of sum and product is:

$$\boxed{\frac{\alpha+\beta}{(\alpha\beta)^{5/8}}}$$

Step 3: Use the quadratic equation

The roots $$\alpha, \beta$$ of the equation:

$$x^2 - 64x + 256 = 0$$

give, by Vieta's formulas:

$$\boxed{\alpha+\beta = 64}$$

and

$$\boxed{\alpha\beta = 256}$$

Substitute these values into the simplified expression:

$$\frac{\alpha+\beta}{(\alpha\beta)^{5/8}} = \frac{64}{256^{5/8}}$$

Since $256 = 2^8$:

$$256^{5/8} = (2^8)^{5/8} = 2^5 = 32$$

Hence:

$$\frac{64}{32} = \boxed{2}$$

Final Answer

$$\boxed{2}$$