6 Given complex numbers $z_{1}$ and $z_{2}$ satisfy $\left|z_{2}\right|=4,4 z_{1}^{2}-2 z_{1} z_{2}+z_{2}^{2}=0$, then the maximum value of $\left|\left(z_{1}+1\right)^{2}\left(z_{1}-2\right)\right|$ is $\qquad$ .
(6) $6 \sqrt{6}$ Hint: Given $\left(z_{1}-z_{2}\right)^{2}=-3 z^{2}$, then $z_{1}-z_{2}= \pm \sqrt{3} \mathrm{i} z_{1}$, $z_{2}=z_{1}(1 \pm \sqrt{3} \mathrm{i})$, so $\left|z_{2}\right|=\left|z_{1}\right| \cdot 2$, since $\left|z_{2}\right|=4$, then $\left|z_{1}\right|=2$. Let $z_{1}=2(\cos \theta+\mathrm{i} \sin \theta)$.
$$
\begin{aligned}
M & =\left|\left(z_{1}+1\right)^{2}\left(z_{1}-2\right)\right|=\left|z_{1}+1\right|^{2} \cdot\left|z_{1}-2\right| \\
& =\left[(2 \cos \theta+1)^{2}+(2 \sin \theta)^{2}\right] \cdot \sqrt{(2 \cos \theta-2)^{2}+(2 \sin \theta)^{2}} \\
& =\sqrt{(5+4 \cos \theta)^{2}(8-8 \cos \theta)} \\
& \leq \sqrt{\left(\frac{(5+4 \cos \theta)+(5+4 \cos \theta)+(8-8 \cos \theta)}{3}\right)^{3}} \\
& =6 \sqrt{6} .
\end{aligned}
$$

When and only when $5+4 \cos \theta=8-8 \cos \theta$, i.e., $\cos \theta=\frac{1}{4}$, $M$ reaches its maximum value $6 \sqrt{6}$.