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:

We have to evaluate the expression

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

where $$\alpha$$ and $$\beta$$ are the roots of the quadratic equation

$$x^{2}-64x+256=0.$$

For every quadratic equation $$x^{2}-Sx+P=0,$$ the sum and product of its roots are given by Vieta’s relations:

$$\alpha+\beta=S,\qquad \alpha\beta=P.$$

Comparing with our equation, we obtain

$$\alpha+\beta=64,\qquad \alpha\beta=256.$$

Now we begin to mould the required expression. Observe that

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

Hence the complete expression can be rewritten as

$$E=\alpha^{3/8}\beta^{-5/8}+\beta^{3/8}\alpha^{-5/8}.$$

To simplify, we introduce the ratio

$$t=\frac{\alpha}{\beta}\quad\Longrightarrow\quad\frac{\beta}{\alpha}=\frac1t.$$

Because $$\alpha\beta=256,$$ we may write

$$\alpha=16\sqrt{t},\qquad \beta=\frac{16}{\sqrt{t}}.$$

Substituting these forms into each term of $$E:$$

First term:

$$\alpha^{3/8}\beta^{-5/8}= \left(16\sqrt{t}\right)^{3/8}\left(\frac{16}{\sqrt{t}}\right)^{-5/8} =16^{-1/4}\,t^{3/16}\,t^{5/16} =16^{-1/4}\,t^{1/2}.$$

Second term:

$$\beta^{3/8}\alpha^{-5/8}= \left(\frac{16}{\sqrt{t}}\right)^{3/8}\left(16\sqrt{t}\right)^{-5/8} =16^{-1/4}\,t^{-3/16}\,t^{-5/16} =16^{-1/4}\,t^{-1/2}.$$

Combining both terms gives

$$E=16^{-1/4}\left(t^{1/2}+t^{-1/2}\right).$$

Because $$16=2^{4},$$ we have $$16^{-1/4}=2^{-1}=\dfrac12,$$ so

$$E=\frac12\left(\sqrt{t}+\frac1{\sqrt{t}}\right).$$

The task has now been reduced to finding $$\sqrt{t}+\dfrac1{\sqrt{t}}.$$ To do that, we need the exact value of $$t=\dfrac{\alpha}{\beta}.$$ The two roots of the given quadratic are

$$\alpha=\frac{64+\sqrt{64^{2}-4\cdot256}}{2}=32+16\sqrt3,$$

$$\beta=\frac{64-\sqrt{64^{2}-4\cdot256}}{2}=32-16\sqrt3.$$

Therefore

$$t=\frac{\alpha}{\beta}=\frac{32+16\sqrt3}{32-16\sqrt3} =\frac{2+\sqrt3}{2-\sqrt3}.$$

Multiplying numerator and denominator by the conjugate $$2+\sqrt3,$$ we get

$$t=\frac{(2+\sqrt3)^2}{(2+\sqrt3)(2-\sqrt3)} =\frac{7+4\sqrt3}{1}=7+4\sqrt3.$$

Notice that

$$(2+\sqrt3)^2=7+4\sqrt3,$$

which implies

$$\sqrt{7+4\sqrt3}=2+\sqrt3.$$

Hence

$$\sqrt{t}=2+\sqrt3,\qquad \frac1{\sqrt{t}}=\frac1{2+\sqrt3}=2-\sqrt3$$

(because $$(2+\sqrt3)(2-\sqrt3)=1$$).

Thus

$$\sqrt{t}+\frac1{\sqrt{t}}=(2+\sqrt3)+(2-\sqrt3)=4.$$

Finally, substituting this sum back into $$E$$ gives

$$E=\frac12\left(4\right)=2.$$

Hence, the correct answer is Option A.