In a triangle $ABC$, if a is the arithmetic mean and $b,c(b\ne c)$  are the two geometric means between any two positive real numbers, then $\frac{{\sin }^{3}B+{\sin }^{3}C}{\sin A\sin B\sin C}=$

Let x and y be two real given positive numbers so $a=\frac{x+y}{2}$ and $b=xr$

$C=x{r}^2$ and $y=x{r}^3$

Now $\frac{{\sin }^{3}B+{\sin }^{3}C}{\sin A+\sin B+\sin C}=\frac{b^3+c^3}{abc}$

$\begin{array}{l}=\frac{{\left(xr\right)}^3+{\left(x{r}^2\right)}^3}{\frac{x+x{r}^3}{2}.xr.x{r}^2}\\ =\frac{x^3{r}^3+x^3{r}^6}{x^3{r}^3\left(\frac{1+r^3}{2}\right)}\end{array}$

$\begin{array}{l}=\frac{2r^3(1+r^3)}{r^3(1+r^3)}\\ =2\end{array}$