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.

Hint:  Note this indenty
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$$
so
$$(\sin{x}+\sin{(2x)})(\sin{(2x)}+\sin{(3x)})(\sin{x}+\sin{(3x)})=0$$
then
$$\sin{x}\cos{x}(2\cos{x}+1)(4\cos^2{x}+2\cos{x}-1)=0$$
then you can solve it
