Show that $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $z=-\frac12(1\pm i\cot(k\pi/8))$; and follow-up questions 
*

*(a) Show that the equation $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $-\frac12\left(1\pm i\cot\frac{k\pi}{8}\right)$, where $k=1,2,3$.

*(b) Hence show that
$$(z+1)^8-z^8=\tfrac18(2z+1)(2z^2+2z+1)\left(4z^2+4z+\csc^2\tfrac{\pi}{8}\right)\left(4z^2+4z+\csc^2\tfrac{3\pi}{8}\right)$$

*(c) By making a suitable substitution into this identity, deduce that
$$\cos^{16}\theta - \sin^{16}\theta = \tfrac1{16}\cos 2\theta(\cos^22\theta+1)\left(\cos^22\theta+\cot^2\tfrac{\pi}{8}\right)\left(\cos^22\theta+\cot^2\tfrac{3\pi}{8}\right) $$
original problem image

How do you reformat the polynomial to reach the answer required?

I have tried to turn the LHS into $[(z+1)/z]^8 = 0$ and solve the roots of unity, but I keep getting $\operatorname{cis}(2k\pi/8)$ instead of $k\pi/8$ which is required. 
Thanks in advance
 A: By your method
$$(z+1)^8-z^8=0 \iff \left(1+\frac1z\right)^8=1$$
then let
$$w=1+\frac1z$$
and solve
$$w^8=1 \implies w=e^{\frac{2i\pi k}8}\quad k=0,1,2,3,4,5,6,7$$
but only $7$ solutions are valid since $k=0$ is not acceptable, indeed the original equation is of degree $7$.
The solutions are given by
$$z=\frac1{e^{\frac{2i\pi k}8}-1}=\frac{e^{\frac{-i\pi k}8}}{e^{\frac{i\pi k}8}-e^{\frac{-i\pi k}8}}=\frac{\cos\left(\frac{\pi k}8\right)-i\sin\left(\frac{\pi k}8\right)}{2i\sin\left(\frac{\pi k}8\right)}=-\frac12-\frac i2\cot \left(\frac{\pi k}8\right)$$
A: $$
(1+z)^8=z^8\tag1
$$
$(1)$ implies that
$$
1+z=e^{\pi ik/4}z\tag2
$$
That is
$$
\begin{align}
z
&=\frac1{e^{\pi ik/4}-1}\\
&=\frac1{\cos(\pi k/4)-1+i\sin(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{(\cos(\pi k/4)-1)^2+\sin^2(\pi k/4)}\\
&=\frac{\cos(\pi k/4)-1-i\sin(\pi k/4)}{2-2\cos(\pi k/4)}\\
&=\frac{-\sin^2(\pi k/4)-i\sin(\pi k/4)(1+\cos(\pi k/4)}{2\sin^2(\pi k/4))}\\
&=-\frac12-\frac i2\cot(\pi k/8)\tag3
\end{align}
$$
where we have used this identity
$$
\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}\tag4
$$
