If $\alpha $ and $\beta $ be two roots of the equation x² – 64x + 256 = 0. Then the value of${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$ is :

$ x^2 - 64x + 256 = 0 $ $ \alpha + \beta = 64,\ \alpha \beta = 256 $ Roots: $ x = \frac{64 \pm \sqrt{64^2 - 4 \cdot 256}}{2} = \frac{64 \pm \sqrt{4096 - 1024}}{2} = \frac{64 \pm \sqrt{3072}}{2} = 32 \pm 16\sqrt{3} $ So, $ \alpha = 32 + 16\sqrt{3},\ \beta = 32 - 16\sqrt{3} $ Now, $ \left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8} $ $ = \frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}} $ $ = \left(\frac{\alpha}{\beta}\right)^{3/8} \cdot \frac{1}{\beta^{1/4}} + \left(\frac{\beta}{\alpha}\right)^{3/8} \cdot \frac{1}{\alpha^{1/4}} $ But use symmetry: Let $ \frac{\alpha}{\beta} = t $ Then $ \frac{\beta}{\alpha} = \frac{1}{t} $ Expression becomes: $ t^{3/8} \cdot t^{-5/8} + t^{-3/8} \cdot t^{5/8} = t^{-1/4} + t^{1/4} $ So, $ = \left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{\alpha}{\beta}\right)^{1/4} $ $ = \sqrt[4]{\frac{\alpha}{\beta}} + \sqrt[4]{\frac{\beta}{\alpha}} $ Now, $ \frac{\alpha}{\beta} = \frac{32+16\sqrt{3}}{32-16\sqrt{3}} = \frac{(2+\sqrt{3})}{(2-\sqrt{3})} = (2+\sqrt{3})^2 $ So, $ \sqrt[4]{\frac{\alpha}{\beta}} = \sqrt{2+\sqrt{3}} $ Similarly, $ \sqrt[4]{\frac{\beta}{\alpha}} = \sqrt{2-\sqrt{3}} $ Hence, $ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 $ But known identity: $ \sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}} = \sqrt{3} + 1 = 2 $