Find the least positive angle satisfying the trigonometric equation $\sin^3 x+\sin^3 2x+\sin^3 3x=(\sin x+\sin 2x+\sin 3x)^3$. I did solve the question, but my method is highly tedious. I combined the sin and then opened the cubic.... Is there some trick? Something I am missing? Thanks.

For $\sin ax+\sin bx=0$
Method $\#1:$
$\sin ax+\sin bx=0\iff\sin ax=-\sin bx=\sin(-bx)$ as $\sin(-A)=-\sin A$
$\implies ax=n\pi+(-1)^n(-bx)$ where $n$ is any integer
If $n$ is even $=2m$(say), $ax+2m\pi-bx\iff x=\dfrac{2m\pi}{a+b}$
Similarly for odd $n=2m+1$(say)
Method $\#2:$
Using Prosthaphaeresis Formula, $$\sin ax+\sin bx=2\sin\dfrac{(a+b)x}2\cos\dfrac{(a-b)x}2$$
If $\sin\dfrac{(a+b)x}2=0, \dfrac{(a+b)x}2=p\pi$
and if $\cos\dfrac{(a-b)x}2=0,\dfrac{(a-b)x}2=\dfrac{(2q+1)\pi}2$  where $p,q$ are integers
