2. Let $\left|z_{1}\right|=\left|z_{2}\right|=a(a \neq 0)$, and $z_{1}+z_{2}$ $=m+m \mathrm{i}$, where $m$ is a non-zero real number. Then the value of $z_{1}^{3} \cdot z_{2}^{3}$ is $\qquad$ .
2. $-a^{6} \mathrm{i}$.

Let $z_{1}=a(\cos \alpha+i \sin \alpha), z_{2}=(\cos \beta+i \sin \beta)$.
Then according to the given conditions, we have
$$
\left\{\begin{array}{l}
\cos \alpha+\cos \beta=\frac{m}{a}, \\
\sin \alpha+\sin \beta=\frac{m}{a},
\end{array}\right.
$$

which means
$$
\left\{\begin{array}{l}
2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=\frac{m}{a}, \\
2 \sin \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=\frac{m}{a}.
\end{array}\right.
$$
Therefore, $\tan \frac{\alpha+\beta}{2}=1, \alpha+\beta=2 k \pi+\frac{\pi}{2}$.
Thus, $z_{1} \cdot z_{2}=a^{2} \mathrm{i},\left(z_{1} \cdot z_{2}\right)^{3}=-a^{6} \mathrm{i}$.