Question

If ${(4+\sqrt{15})}^{3/2}-{(4-\sqrt{15})}^{3/2}=k\sqrt{6}$, then $k=$............

• A. $9$
• B. $8$
• C. $7$
• D. $6$

Answer 1

The correct option is A $9$

$4+\sqrt{15}={\left(\frac{\sqrt{5}}{2}\right)}^2+2\frac{\sqrt{5}}{\sqrt{2}}\times \frac{\sqrt{3}}{\sqrt{2}}+{\left(\sqrt{\frac{3}{2}}\right)}^2$

$4+\sqrt{15}={\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\right)}^2$

similarly $4-\sqrt{15}={\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\right)}^2$

As $(4+\sqrt{15}{)}^{3/2}-(4-\sqrt{15}{)}^{3/2}=k\sqrt{6}$

$\Rightarrow \frac{(\sqrt{5}+\sqrt{3}{)}^3-(\sqrt{5}-\sqrt{3}{)}^3}{(\sqrt{2}{)}^3}=\frac{2(\sqrt{3}{)}^3+6(\sqrt{5}{)}^2\left(\sqrt{3}\right)}{2\sqrt{2}}$

$\Rightarrow \frac{3\sqrt{3}+15\sqrt{3}}{\sqrt{2}}=9\sqrt{6}\Rightarrow k=9$