{frac{{[4 + sqrt{15}]}^{3/2} + {[4 - sqrt{15} | vedclass

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(D) Let $a = 4 + \sqrt{15}$ and $b = 4 - \sqrt{15}$. Note that $a \cdot b = 16 - 15 = 1$. Thus $b = 1/a$.Similarly,let $c = 6 + \sqrt{35}$ and $d = 6

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None of these

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(D) Let $a = 4 + \sqrt{15}$ and $b = 4 - \sqrt{15}$. Note that $a \cdot b = 16 - 15 = 1$. Thus $b = 1/a$.Similarly,let $c = 6 + \sqrt{35}$ and $d = 6 - \sqrt{35}$. Note that $c \cdot d = 36 - 35 = 1$. Thus $d = 1/c$.The expression is $E = \frac{a^{3/2} + b^{3/2}}{c^{3/2} - d^{3/2}} = \frac{a^{3/2} + a^{-3/2}}{c^{3/2} - c^{-3/2}}$.We know that $4 + \sqrt{15} = \frac{8 + 2\sqrt{15}}{2} = \frac{(\sqrt{5} + \sqrt{3})^2}{2}$.So $a^{3/2} = \frac{(\sqrt{5} + \sqrt{3})^3}{2\sqrt{2}}$.Similarly,$6 + \sqrt{35} = \frac{12 + 2\sqrt{35}}{2} = \frac{(\sqrt{7} + \sqrt{5})^2}{2}$.So $c^{3/2} = \frac{(\sqrt{7} + \sqrt{5})^3}{2\sqrt{2}}$.Substituting these into the expression:$a^{3/2} + a^{-3/2} = \frac{(\sqrt{5} + \sqrt{3})^3 + (\sqrt{5} - \sqrt{3})^3}{2\sqrt{2}} = \frac{2(5\sqrt{5} + 3\sqrt{5})}{2\sqrt{2}} = \frac{8\sqrt{5}}{2\sqrt{2}} = 2\sqrt{10}$.$c^{3/2} - c^{-3/2} = \frac{(\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3}{2\sqrt{2}} = \frac{2(3 \cdot 7\sqrt{5} + 5\sqrt{5})}{2\sqrt{2}} = \frac{26\sqrt{5}}{2\sqrt{2}} = \frac{13\sqrt{5}}{\sqrt{2}} = \frac{13\sqrt{10}}{2}$.$E = \frac{2\sqrt{10}}{13\sqrt{10}/2} = \frac{4}{13}$.Wait,re-evaluating the simplification: $a^{3/2} + b^{3/2} = \frac{(\sqrt{5}+\sqrt{3})^3 + (\sqrt{5}-\sqrt{3})^3}{2\sqrt{2}} = \frac{2(5\sqrt{5} + 3\sqrt{5})}{2\sqrt{2}} = \frac{8\sqrt{5}}{2\sqrt{2}} = 2\sqrt{10}$.$c^{3/2} - d^{3/2} = \frac{(\sqrt{7}+\sqrt{5})^3 - (\sqrt{7}-\sqrt{5})^3}{2\sqrt{2}} = \frac{2(3 \cdot 7\sqrt{5} + 5\sqrt{5})}{2\sqrt{2}} = \frac{26\sqrt{5}}{2\sqrt{2}} = \frac{13\sqrt{5}}{\sqrt{2}} = \frac{13\sqrt{10}}{2}$.$E = \frac{2\sqrt{10}}{13\sqrt{10}/2} = 4/13$. Since $4/13$ is not in the options,the correct choice is $D$.

${\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}} = $

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