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Question Number 96171 by bemath last updated on 30/May/20

$${(4+\sqrt{15})}^{\frac{3}{2}}-{(4-\sqrt{15})}^{\frac{3}{2}}=\mathrm{k}\sqrt{6}$$$$\mathrm{find}\mathrm{k}$$

Commented by bobhans last updated on 30/May/20

$$\mathrm{let}\sqrt{4+\sqrt{15}}=\mathrm{u}\mathrm{\&}\sqrt{4-\sqrt{15}}=\mathrm{v}$$$${\mathrm{u}}^3-{\mathrm{v}}^3=(\mathrm{u}-\mathrm{v})({\mathrm{u}}^2+\mathrm{uv}+{\mathrm{v}}^2)=\mathrm{k}\sqrt{6}$$$$(\mathrm{u}-\mathrm{v})(8+1)=\mathrm{k}\sqrt{6}$$$$\mathrm{u}-\mathrm{v}=\frac{\mathrm{k}\sqrt{6}}{9}\Rightarrow {\mathrm{u}}^2-2\mathrm{uv}+{\mathrm{v}}^2=\frac{{6\mathrm{k}}^2}{81}$$$$8-2=\frac{{6\mathrm{k}}^2}{81}\Rightarrow \mathrm{k}=9.$$$$\mathrm{since}{\mathrm{u}}^3-{\mathrm{v}}^3\mathrm{is}\mathrm{positive}$$

Commented by bemath last updated on 30/May/20

$$\mathrm{thanks}$$

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