No, for example pick $\zeta = \exp \frac{i\pi}{15}$, a primitive $30$th root of unity.
Then $0 = (1 + \zeta^{10} + \zeta^{20}) + \zeta^{15}(1 + \zeta^6 + \zeta^{12} + \zeta^{18} + \zeta^{24}) - (1 + \zeta^{15}) \\
= \zeta^3 + \zeta^9 + \zeta^{10} + \zeta^{20} + \zeta^{21} + \zeta^{27}$
But you can't partition those $6$ roots into vertices of regular $k$-gons.
However, it is true that you can always add vertices of regular $k$-gons to your set of roots (possibly adding the same root multiple times) to obtain a new multiset that you can then arrange into a sum of vertices of regular $k$-gons (possibly using the same $k$-gon multiple times).
Let $\zeta_n = \exp{\frac{2i\pi}n}$ and $f$ be the map $\Bbb Z^n \to \Bbb Z[\zeta_n]$ given by $(a_i) \mapsto \sum a_i \zeta_n^i$.
The minimal polynomial of $\zeta_n$ over $\Bbb Q$ is the cyclotomic polynomial $\Phi_n$, of degree $\varphi(n)$. Hence $\Bbb Z[\zeta_n]$ is free of rank $\varphi(n)$.
If $n = \prod p_i^{d_i}$, then $\zeta$ is a primitive $n$th root of unity iff it is of order exactly $n$, and not $n/p$ for any $p$, iff $\zeta^{n/p}$ is a primitive $p$th root, for all $p$. Hence $\Phi_n$ is the gcd of the $\Phi_p(X^{n/p}) = 1 + X^{n/p} + X^{2n/p} + \ldots + X^{(p-1)n/p}$, and $\ker f$ is generated by those relations, which correspond to vertices of regular $p$-gons.
Note that when $Z^3 - 1$ is factored as $(Z-1)(Z^2+Z+1)$, it reduces the problem of solving $Z^3-1$ into two subproblems: Either $Z-1=0$ or $Z^2+Z+1=0$.
The first of these gives us $Z=1$ as a solution, and the second of these yields, by the quadratic formula, a pair of conjugate complex roots:
$$ Z = \frac{-1 \pm \sqrt{-3}}{2} = -\frac{1}{2} \pm i \frac{\sqrt{3}}{2} $$
These two conjugate complex roots fall in the second and third quadrants. The one in the second quadrant has the smallest positive "argument" (angle with respect to the positive real axis).
You then say you "need this" to simplify $(1+w^2)(1+w)$. That part of the Question is unclear. Do you mean that $w$ is one of the three roots of $Z^3-1=0$? Do you mean that it is specifically the one with "smallest positive argument"?
If you mean the latter, then you can "plug in" the root (found above, with the plus sign for the choice of plus/minus)) and get a complex number. If the former, you can do a bit of algebra, since $(1+w^2)(1+w) = 1 + w + w^2 + w^3$. That is:
$$ 1+w+w^2+w^3 = 0+w^3 = 1 $$
for either of the complex roots of $Z^3 - 1 = 0$.
Best Answer
The right hand side should be
$$\frac{1-e^{\frac{in2\pi}{n}}}{1-e^{\frac{i2\pi}{n}}}=0$$
since $e^{\frac{in2\pi}{n}}=e^{i2\pi}=1$.