Question

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

• A. $9$
• B. $8$
• C. $7$
• D. $6$

Answer 1

The correct option is A $9$

As $4+\sqrt{15}=\frac{1}{2}(8+2\sqrt{15})=\frac{1}{2}(\sqrt{5}+\sqrt{3}{)}^2$

$\Rightarrow (4+\sqrt{15}{)}^{3/2}-(4-\sqrt{15}{)}^{3/2}=\frac{1}{2\sqrt{2}}(\sqrt{5}+\sqrt{3}{)}^3-\frac{1}{2\sqrt{2}}(\sqrt{5}-\sqrt{3}{)}^3$

$=\frac{1}{2\sqrt{2}}(a^3-b^3)=\frac{1}{2\sqrt{2}}(a-b)(a^2+b^2+ab)=\frac{1}{2\sqrt{2}}\left(2\sqrt{3}\right)\left(\right(a+b{)}^2-ab)$

$=\frac{\sqrt{3}}{\sqrt{2}}(4\times 5-(5-3\left)\right)=\frac{\sqrt{3}}{\sqrt{2}}\left(18\right)=k\sqrt{6}\Rightarrow k=9$

$\Rightarrow \left(A\right)$