Cute cube roots

So basically we need to find this summation:

$\dfrac{1}{\omega} + \dfrac{1}{\omega^2}+ \dfrac{1}{\omega^3} + \dots + \dfrac{1}{\omega^{1001}}$

Note that

$\dfrac{1}{\omega}=\dfrac{\omega^2}{\omega^3}=\omega^2 \ , \ \dfrac{1}{\omega^2}=\dfrac{\omega}{\omega^3}=\omega \ , \ \dfrac{1}{\omega^3}=1 \\ \Rightarrow \dfrac{1}{\omega} + \dfrac{1}{\omega^2}+ \dfrac{1}{\omega^3}= \omega^2+\omega+1=0$

Thus we have $\dfrac{1}{\omega} + \dfrac{1}{\omega^2}+ \dfrac{1}{\omega^3}+ \dots + \dfrac{1}{\omega^{999}} = 0+0+0+\dots + 0 =0$

So our problem reduces to:

$\sum_{i=1}^{1001} \dfrac{1}{\omega^i}=\dfrac{1}{\omega^{1000}}+\dfrac{1}{\omega^{1001}} = \dfrac{1}{\omega}+\dfrac{1}{\omega^2} = \omega^2+\omega=\boxed{-1} \because 1+\omega+\omega^2=0$

Alternatively , we can also use formula for geometric progression and get our desired result.

If we denote the sum by $S$ , then $1+S=\sum_{i=0}^{1001}\frac{1}{\omega^i}=\frac{1-\frac{1}{\omega^{1002}}}{1-\frac{1}{\omega}}=\frac{1-\frac{1}{(\omega^3)^{334}}}{1-\frac{1}{\omega}}=0$ since $\omega^3=1$ so $S=\boxed{-1}$