8.216. $\sin 3z + \sin^3 z = \frac{3 \sqrt{3}}{4} \sin 2z$.
## Solution.

Since $\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha$ and $\sin 2 \alpha=2 \sin \alpha \cos \alpha$, we have:

$3 \sin z-4 \sin ^{3} z+\sin 3 z-\frac{3 \sqrt{3}}{2} \sin z \cos z=0 \Leftrightarrow$

$\Leftrightarrow 3 \sin z-3 \sin ^{3} z-\frac{3 \sqrt{3}}{2} \sin z \cos z=0 \Leftrightarrow$

$\Leftrightarrow 3 \sin z\left(1-\sin ^{2} z-\frac{\sqrt{3}}{2} \cos z\right)=0 \Leftrightarrow \sin z\left(\cos ^{2} z-\frac{\sqrt{3}}{2} \cos z\right)=0 \Leftrightarrow$ $\Leftrightarrow \sin z \cos z\left(\cos z-\frac{\sqrt{3}}{2}\right)=0$.

From this,

1) $\sin z=0, z_{1}=\pi k, k \in \mathbb{Z};$
2) $\cos z=0, z_{2}=\frac{\pi}{2}+\pi n=\frac{\pi}{2}(2 n+1), n \in \mathbb{Z}$;
3) $\cos z-\frac{\sqrt{3}}{2}=0, \cos z=\frac{\sqrt{3}}{2}, z_{3}= \pm \frac{\pi}{6}+2 \pi l, l \in \mathbb{Z}$.

Answer: $z_{1}=\pi k ; z_{2}=\frac{\pi}{2}(2 n+1) ; z_{3}= \pm \frac{\pi}{6}+2 \pi l, k, n$ and $l \in \mathbb{Z}$.