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 $n=-1,0,1$

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

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

Answer 2

$$\begin{align}\sin x\cdot\sin\left(\dfrac{\pi}{3}+x\right)\cdot\sin\left(\dfrac{\pi}{3}-x\right)&=\sin x\left(\sin^2\dfrac{\pi}{3}-\sin^2 x\right)\\ &=\sin x\left(\left(\dfrac{\sqrt{3}}{2}\right)^2-\sin^2 x\right)\\ &=\sin x\left(\dfrac{3-4\sin^2x}{4}\right)\\ &=\dfrac{3\sin x-4\sin^3 x}{4}\\ &=\dfrac{1}{4}\sin 3x\end{align}$$ Therefore, $\sin x\cdot\sin\left(\dfrac{\pi}{3}+x\right)\cdot\sin\left(\dfrac{\pi}{3}-x\right)=\dfrac{1}{4}\sin 3x$