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

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(D) Let $x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$. Then $x+y = 8$ and $xy = 16 - 15 = 1$. We can write $x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqr

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(D) Let $x = 4 + \sqrt{15}$ and $y = 4 - \sqrt{15}$. Then $x+y = 8$ and $xy = 16 - 15 = 1$. We can write $x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2x} + \sqrt{2y})^3$. Since $2x = 8 + 2\sqrt{15} = (\sqrt{5} + \sqrt{3})^2$ and $2y = (\sqrt{5} - \sqrt{3})^2$,we have $\sqrt{2x} = \sqrt{5} + \sqrt{3}$ and $\sqrt{2y} = \sqrt{5} - \sqrt{3}$. Thus,$x^{3/2} + y^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{5} + \sqrt{3} + \sqrt{5} - \sqrt{3})^3 = \frac{1}{\sqrt{2}} (2\sqrt{5})^3 = \frac{40\sqrt{5}}{\sqrt{2}} = 20\sqrt{10}$. Similarly,for the denominator,let $a = 6 + \sqrt{35}$ and $b = 6 - \sqrt{35}$. Then $2a = 12 + 2\sqrt{35} = (\sqrt{7} + \sqrt{5})^2$ and $2b = (\sqrt{7} - \sqrt{5})^2$. $a^{3/2} - b^{3/2} = \frac{1}{\sqrt{2}} (\sqrt{2a}^3 - \sqrt{2b}^3) = \frac{1}{\sqrt{2}} ((\sqrt{7} + \sqrt{5})^3 - (\sqrt{7} - \sqrt{5})^3)$. Using $(u+v)^3 - (u-v)^3 = 2v^3 + 6u^2v$,we get $\frac{1}{\sqrt{2}} (2(5\sqrt{5}) + 6(7)\sqrt{5}) = \frac{1}{\sqrt{2}} (10\sqrt{5} + 42\sqrt{5}) = \frac{52\sqrt{5}}{\sqrt{2}} = 26\sqrt{10}$. The ratio is $\frac{20\sqrt{10}}{26\sqrt{10}} = \frac{20}{26} = \frac{10}{13}$.

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

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