Unity in Math

$\large \displaystyle z_1 + z_2 + z_3 = 2(1 + \omega + \omega ^2) + 3(1 + \omega + \omega ^2) = 0\\ \large \displaystyle \therefore z_1^3 + z_2^3 + z_3^3 = 3 z_1 z_2 z_3 = 15[4 + 9 + 6(\omega + \omega ^2)]\\ \large \displaystyle = 15 ( 4 + 9 - 6) = 15 \times 7 = \color{#D61F06}{\boxed{105}}.$

Lengthier method.

Using the fact that $\omega^3=1$ and $\omega+\omega^2=-1$ we have $z_1=2\omega+3\omega^2=-2+\dfrac 1 {\omega}$ . $z_2=2\omega^2+3\omega=-2+\omega$ .

Now expand! We have $z_1^3+z_2^3+z_3^3=-8+ \dfrac 1 {\omega^3}+\dfrac {12}{\omega}-\dfrac 6 {\omega^2}-8+1+12\omega-6\omega^2+125$ .

Simplifying we get the value as $\boxed{105}$ .