roots of complex polynomial - tricks What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$

$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to
$$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$$
So now we need to get the values from t for $k = 0,...5$.
How to get the values of t from the following identity then?
$$
\begin{align}
\frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
(t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\
1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\
\end{align}
$$
And now?
$$
t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1}
$$
So I've got six roots for $k = 0,...5$ as follows
$$
t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1}
$$
Is this right? But how can it be that the bottom equals $0$ for $k=0$?
I don't exactly know how to simplify this:
$$\frac{  \frac{1}{  e^{i \cdot k \cdot \frac{2 \pi}{6}}  } + 1  }{ 1 - \frac{1}{  e^{i \cdot k \cdot \frac{2 \pi}{6}}  }}$$

Note that
$$(t+1)^6 - (t-1)^6=((t+1)^3-(t-1)^3)((t+1)^3+(t-1)^3)$$
(difference of squares).
When you simplify the first term in the product on the right, there is no $t^3$ term and no $t$ term!  The second term in the product simplifies to $2t^3+6t$.
Remark: The solution by Arhabhata is the right one, it works if we replace $6$ by $n$. And when we set $\frac{t-1}{t+1}=e^{2\pi i k/n}$, where $k=1,2,\dots,n-1$,  and solve for $t$, we get $-i$ times cotangents. 
