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\((4 + \sqrt{15})^\frac {3}{2} + (4 - \sqrt{15}) ^\frac{3}{2}= k\sqrt10 \)হলে, k এর মান 7

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\((4 + \sqrt{15})^\frac {3}{2}\)
\(=(\cfrac{8}{2}+ \sqrt{15})^\frac {3}{2}\)
\(=(\cfrac{5+3}{2}+ \sqrt{15})^\frac {3}{2}\)
\(=(\cfrac{5}{2}+ \cfrac{3}{2}+\sqrt{15})^\frac {3}{2}\)
\(=\left[(\cfrac{\sqrt5}{\sqrt2})^2+ (\cfrac{\sqrt3}{\sqrt2})^2+2.\cfrac{\sqrt5}{\sqrt2}.\cfrac{\sqrt3}{\sqrt2}\right]^\frac {3}{2}\)
\(=\left[\cfrac{\sqrt5}{\sqrt2}+\cfrac{\sqrt3}{\sqrt2}\right]^{2\times \frac{3}{2}}\)
\(=\left[\cfrac{\sqrt5}{\sqrt2}+\cfrac{\sqrt3}{\sqrt2}\right]^3\)

অনুরূপভাবে, \((4 - \sqrt{15})^\frac {3}{2}=\left[\cfrac{\sqrt5}{\sqrt2}-\cfrac{\sqrt3}{\sqrt2}\right]^3\)

এখন ধরি, \(\cfrac{\sqrt5}{\sqrt2}=a\) এবং \(\cfrac{\sqrt3}{\sqrt2}=b\)
\(\therefore (a+b)^3+(a-b)^3=k\sqrt{10}\)
বা, \(a^3+\cancel{3a^2b}+3ab^2+\cancel{b^3}\) \(+a^3-\cancel{3a^2b}+3ab^2-\cancel{b^3}=k\sqrt{10}\)
বা, \(2a^3+6ab^2=k\sqrt{10}\)
বা, \(2a(a^2+3b^2)=k\sqrt{10}\)
বা, \(2\times \cfrac{\sqrt5}{\sqrt2}\left[(\cfrac{\sqrt5}{\sqrt2})^2+3(\cfrac{\sqrt3}{\sqrt2})^2\right]=k\sqrt{10}\) [\(a, b\) এর মান বসিয়ে ]
বা, \(2\times \cfrac{\sqrt5\times \sqrt 2}{\sqrt2\times \sqrt2}\left[\cfrac{5}{2}+3\times\cfrac{3}{2}\right]=k\sqrt{10}\)
বা, \(2\times \cfrac{\sqrt{10}}{2}\times \cfrac{5+9}{2}=k\sqrt{10}\)
বা, \(\cancel{2}\times \cfrac{\sqrt{10}}{\cancel{2}}\times \cfrac{\cancel{14}7}{\cancel{2}}=k\sqrt{10}\)
বা, \(7\sqrt{10}=k\sqrt{10}\)
বা, \(k=7\)