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(D) Given the quadratic equation $x^{2}-64x+256=0$.From the properties of roots,the sum of roots $\alpha+\beta = 64$ and the product of roots $\alpha\

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$2$

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(D) Given the quadratic equation $x^{2}-64x+256=0$.From the properties of roots,the sum of roots $\alpha+\beta = 64$ and the product of roots $\alpha\beta = 256$.We need to evaluate the expression $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}$.This simplifies to $\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}}$.Taking the common denominator,we get $\frac{\alpha^{3/8} \cdot \alpha^{5/8} + \beta^{3/8} \cdot \beta^{5/8}}{(\alpha\beta)^{5/8}}$.Since $\alpha^{3/8} \cdot \alpha^{5/8} = \alpha^{(3+5)/8} = \alpha^1 = \alpha$,the expression becomes $\frac{\alpha+\beta}{(\alpha\beta)^{5/8}}$.Substituting the values $\alpha+\beta = 64$ and $\alpha\beta = 256 = 2^8$,we get $\frac{64}{(2^8)^{5/8}}$.This simplifies to $\frac{64}{2^5} = \frac{64}{32} = 2$.

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

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