2016/087) If $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$ prove that $ab+bc+ca=0$

Let $asin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})=k$
hence $\frac{k}{a} = \sin\,x\cdots(1)$
$\frac{k}{b} = \sin(x+\frac{2\pi}{3})\cdots(2)$
$\frac{k}{c} = \sin(x+\frac{4\pi}{3})$
or $\frac{k}{c} = \sin(x-\frac{2\pi}{3})\cdots(3)$

from (1),(2) and (3)
$\frac{k}{a} + \frac{k}{b} + \frac{k}{c} =\sin\,x +  \sin(x+\frac{2\pi}{3}) + \sin(x-\frac{2\pi}{3})$
$= \sin\,x +  \sin\,x\cos \frac{2\pi}{3} + \cos \,x\sin  \frac{2\pi}{3} + \sin\,x\cos \frac{2\pi}{3} - \cos \,x\sin  \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x\cos \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x( -\frac{1}{2})$
$= \sin\,x -  \sin\,x$
$= 0$