Prove that $\sin(x) \sin(\frac{\pi}{3}+x)\sin(\frac{\pi}{3}-x)=\frac{1}{4}\sin3x$

Question

Prove that $\sin(x) \sin(\frac{\pi}{3}+x)\sin(\frac{\pi}{3}-x)=\frac{1}{4}\sin3x$
I was able to transform it into $\sin(x)\left(\frac{3}{4}\cos^2x-\frac{1}{4} \sin^2x\right).$ What should I do next?

Answer 1

$$\sin3t=3\sin t-4\sin^3t$$

If $\sin3t=\sin3x,3t=180^\circ n+(-1)^n3x$ where $n$ is any integer

$t=60^\circ n+(-1)^nx$ where $n=-1,0,1$

So, the roots of $$4\sin^3t-3\sin t+\sin3x=0$$ are $\sin t$ where $t=60^\circ n+(-1)^nx$ where $n=-1,0,1$

$\implies\sin(-60^\circ-x)\sin x\sin(60^\circ-x)=(-1)^3\dfrac{\sin3x}4$

$\iff4\sin(60^\circ+x)\sin x\sin(60^\circ-x)=\sin3x$