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

Given quadratic equation ${\mathrm{x}}^2-64\mathrm{x}+256=0$ having roots $\alpha$and $\beta$ , so

$\mathrm{\alpha }+\mathrm{\beta }=64$ and $\mathrm{\alpha \beta }=256$

now,

${\left(\frac{{\mathrm{\alpha }}^3}{{\mathrm{\beta }}^5}\right)}^{1/8}+{\left(\frac{{\mathrm{\beta }}^3}{{\mathrm{\alpha }}^5}\right)}^{1/8}=\frac{\mathrm{\alpha }+\mathrm{\beta }}{{\left(\mathrm{\alpha \beta }\right)}^{{\displaystyle \frac{5}{8}}}}=\frac{64}{2^5}=\frac{64}{32}=2$