Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$ 
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.

I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate thrice but it's very lenghty.

Strigthforward approach:
First we combine $\frac{1}{(\omega^{k}-1)^3}$ with $\frac{1}{(\omega^{15-k}-1)^3}$ to get rid of $i$ in the denominator.
$$\sum\limits_{k=1}^{14}\frac{1}{(\omega^k-1)^3}=
\sum\limits_{k=1}^{7}\frac{(\omega^{15-k}-1)^3+(\omega^{k}-1)^3}
{\left((\omega^k-1)(\omega^{15-k}-1)\right)^3}=$$
$$\sum\limits_{k=1}^{7}\frac{
\left((\omega^{15-k}-1)+(\omega^{k}-1)\right)
\left(
(\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2
\right)
}
{\left(2-\omega^k-\omega^{15-k}\right)^3}=$$
$$-\sum\limits_{k=1}^{7}\frac{
(\omega^{15-k}-1)^2-(\omega^{15-k}-1)(\omega^{k}-1)+(\omega^{k}-1)^2
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$$
$$-\sum\limits_{k=1}^{7}\frac{
\omega^{30-2k}-2\omega^{15-k}+1-2+\omega^k+\omega^{15-k}+\omega^{2k}-2\omega^{k}+1
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$$
$\displaystyle-\sum\limits_{k=1}^{7}\frac{
\omega^{30-2k}+\omega^{2k}-\omega^{15-k}-\omega^{k}
}
{\left(2-\omega^k-\omega^{15-k}\right)^2}=$
$\displaystyle-\sum\limits_{k=1}^{7}\frac{\omega^{2k} + \omega^{k} + 1}{\omega^{2k} - 2 \omega^{k} + 1}
$
$=\displaystyle-7-3\sum\limits_{k=1}^{7}\frac{1}{\omega^{k} - 2 +  \omega^{-k}}=$
$$-7-\frac32\sum\limits_{k=1}^{7}\frac{1}{\cos\frac{2k\pi}{15}-1}=
-7+\frac34\sum\limits_{k=1}^{7}\frac{1}{\sin^2\frac{k\pi}{15}}=$$
with a large overkill
$\displaystyle -7+\frac34\cdot\frac23\cdot 7 \cdot 8=21$

