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.
 A: Let $n = 15$ and $P(x) = \frac{x^n-1}{x-1} = \sum\limits_{k=0}^{n-1} x^k$. Last part of your question suggest you already know:
$$\mathcal{S} \stackrel{def}{=} \sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = - \frac12 \left.\frac{d^2}{dx^2} \frac{P'(x)}{P(x)}\right|_{x=1} = -\frac12\left.\frac{d^3}{dx^3}\log P(x)\right|_{x=1}$$
To evaluate the derivative, change variable to $t = \log x$ and let $D$ be the operator $\frac{d}{dt}$, we have
$$-2\mathcal{S} = \left.\left(x^3\frac{d^3}{d x^3}\right)\log P(x)\right|_{x=1}
= D(D-1)(D-2)\left.\log P(e^t)\right|_{t=0}\tag{*1}
$$
Notice
$$\log P(e^t) = \log\frac{e^{nt} - 1}{e^t-1} = \log n + f(nt) - f(t)
\quad\text{ where }\quad f(t) = \log\frac{e^t - 1}{t}$$
We just need to figure out the Taylor expansion of $f(t)$ up to $O(t^4)$. Since
$$f(t) = \log\left( e^{\frac{t}{2}} \frac{\sinh(\frac{t}{2})}{\frac{t}{2}}\right)
= \frac{t}{2} + \log\left( 1 + \frac{t^2}{3! 2^2} + O(t^4)\right)
= \frac{t}{2} + \frac{t^2}{24} + O(t^4)$$
We have
$$Df(t) = \frac12 + \frac{t}{12} + O(t^3)$$
and hence
$$Df(t)|_{t=0} = \frac12,\quad D^2f(t)|_{t=0} = \frac{1}{12}\quad\text{ and }\quad D^3f(t)|_{t=0} = 0$$
Substitute this in $(*1)$, we get
$$-2\mathcal{S} = \bigg[(D^2 - 3D + 2)D(f(nt) - f(t))\bigg]_{t=0} = -\frac{3}{12}(n^2-1) + \frac{2}{2}(n-1)$$
Similplify this give us
$$\sum_{k=1}^{n-1} \frac{1}{(\omega^k - 1)^3} = \mathcal{S} = \frac{(n-3)(n-1)}{8}$$
For $n = 15$, this reduces to $\displaystyle\;\frac{(15-3)(15-1)}{8} = 21$ as first pointed out by @user64494 in comment.
A: $w_k$ are the roots of $$y^{15}-1=0$$
Now let $p_k=\dfrac1{(w_k-1)^3}, w_k-1=\sqrt[3]{\dfrac1{p_k}}$
Writing $p_k$ as $z$ and $w_k$ is a root of $y^{15}=1$
$$1=\left(1+\sqrt[3]{\dfrac1z}\right)^{15}$$
$$z^5=(1+\sqrt[3]z)^{15}$$
$$\iff - \sum_{r=0}^4z^r\binom{15}{3r}=z^{1/3}\sum_{r=0}^4 z^r\binom{15}{3r+1}+z^{2/3}\sum_{r=0}^4 z^r\binom{15}{3r+2} $$
Now to rationalize take cube in both sides,
$$-\left(\sum_{r=0}^4z^r\binom{15}{3r}\right)^3=z\left(\sum_{r=0}^4 z^r\binom{15}{3r+1}\right)^3+z^2\left(\sum_{r=0}^4 z^r\binom{15}{3r+2}\right)^3+3z\left(- \sum_{r=0}^4z^r\binom{15}{3r}\right)$$
$$\left(\binom{15}{3\cdot4+2}\right)^3z^{4\cdot3+2}+z^{4\cdot3+1}\left(\left(\binom{15}{3\cdot4+1}\right)^3-3\binom{15}{12}\right)+\cdots=0$$
$$\implies\sum_{k=1}^{14}p_k=-\dfrac{\left(\binom{15}2\right)^3-3\binom{15}{12}}{\left(\binom{15}1\right)^3}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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With $\ds{\omega \equiv \exp\pars{2\pi\ic/15}}$:
\begin{align}
\sum_{k = 1}^{14}{1 \over \pars{\omega^{k} - 1}^{3}} & =
2\,\Re\sum_{k = 1}^{7}{1 \over \pars{\expo{2k\pi\ic/15} - 1}^{3}}
\\[5mm] & =
2\,\Re\sum_{k = 1}^{7}{1 \over
\expo{k\pi\ic/5}\pars{\expo{k\pi\ic/15} - \expo{-k\pi\ic/15}}^{3}}
\\[5mm] & =
2\,\Re\sum_{k = 1}^{7}{\expo{-k\pi\ic/5} \over
\bracks{2\ic\sin\pars{{k\pi/15}}}^{3}} =
{1 \over 4}\sum_{k = 1}^{7}\underbrace{{\sin\pars{k\pi/5} \over
\sin^{3}\pars{{k\pi/15}}}}_{\ds{3\cot^{2}\pars{k\pi \over 15} - 1}}
\\ & =
-\,{7 \over 4} + {3 \over 4}
\underbrace{\sum_{k = 1}^{7}\cot^{2}\pars{k\pi \over 15}}
_{\ds{\color{red}{\Large\S}\quad{91 \over 3}}} = \bbx{21}
\end{align}
$\ds{\color{red}{\Large\S}}$: I'm still trying to work out the sum !!!.
A: 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
$\displaystyle -7+\frac34\cdot\frac23\cdot 7 \cdot 8=21$

