1. Find the roots of the quadratic

Given equation: $x^2 - 64x + 256 = 0$

Factorize: $x^2 - 64x + 256 = (x-32)^2 - 32^2 + 256$ But more directly, use the discriminant: $D = (-64)^2 - 4\cdot 1 \cdot 256 = 4096 - 1024 = 3072$

So the roots are $\alpha, \beta = \frac{64 \pm \sqrt{3072}}{2}$ Now, $\sqrt{3072} = \sqrt{1024\cdot 3} = 32\sqrt{3}$ Hence, $\alpha, \beta = \frac{64 \pm 32\sqrt{3}}{2} = 32 \pm 16\sqrt{3}$

So we may take $\alpha = 32+16\sqrt{3}, \quad \beta = 32-16\sqrt{3}$

2. Use the product of roots

From the quadratic, $\alpha\beta = 256$ Also, $\alpha = 16(2+\sqrt{3}), \quad \beta = 16(2-\sqrt{3})$ And since $(2+\sqrt{3})(2-\sqrt{3}) = 1$ we get $\alpha\beta = 16\cdot 16 = 256$

3. Simplify each term

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

Rewrite the first term: $\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} = \left(\frac{\alpha^3\alpha^5}{\beta^5\alpha^5}\right)^{1/8} = \left(\frac{\alpha^8}{(\alpha\beta)^5}\right)^{1/8}$ Since $\alpha\beta=256$, $\left(\frac{\alpha^8}{256^5}\right)^{1/8} = \frac{\alpha}{256^{5/8}}$ Now, $256 = 2^8 \implies 256^{1/8} = 2$ Therefore, $256^{5/8} = (256^{1/8})^5 = 2^5 = 32$ So, $\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} = \frac{\alpha}{32}$

Similarly, $\left(\frac{\beta^3}{\alpha^5}\right)^{1/8} = \frac{\beta}{32}$

4. Add the two terms

Hence the required value is $\frac{\alpha}{32} + \frac{\beta}{32} = \frac{\alpha+\beta}{32}$ But for the quadratic $x^2-64x+256=0$, $\alpha+\beta = 64$ Therefore, $\frac{\alpha+\beta}{32} = \frac{64}{32} = 2$

5. Check the options

The value is: $2$ So the correct option is C.