Question

In the $\Delta ABC$, $a$ is the A.M and $b$ and $c$ are two G.M's between two positive real numbers

then $\frac{{\sin }^{3}B+{\sin }^{3}C}{\sin A\sin B\sin C}$ is?

Answer 1

Let $x$ and $y$ be the two positive numbers
$a=\frac{x+y}{2}$...(1)
Also, given $x,b,c,y$ are in G.P.
Let $k$ be the common ratio of the G.P
So, $b=xk,c=k^{2}x,y=k^{3}x$
Now, $\frac{{\sin }^{3}B+{\sin }^{3}C}{sinA\sin B\sin C}$
$=\frac{K^3{b}^3+K^3{c}^3}{K^{3}abc}$; (since $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ )

$=\frac{b^3+c^3}{abc}$
$=\frac{x^3{k}^3+x^3{k}^6}{(\frac{x+k^{3}x}{2})kx.k^{2}x}$
$=\frac{2k^3(1+k^3)}{(1+k^3)k^3}=2$