I have the following:
$$\left(e^{\omega_0x}+e^{\omega_1x}+e^{\omega_2x}\right)^n$$
where $\omega_k=e^\frac{2ki\pi}{3}$. We can change the above to
$$\left[e^{\omega_0x}+(e^{\omega_1x}+e^{\omega_2x})\right]^n$$ and use the Binomial Theorem to get
$$\sum_{k=0}^n\binom{n}{k}{(e^{\omega_0x}})^k(e^{\omega_1}+e^{\omega_2})^{n-k}$$ and again to obtain $$\sum_{k=0}^n\binom{n}{k}{(e^{\omega_0x}})^k\sum_{j=0}^{n-k}\binom{n-k}{j}{(e^{\omega_1x})^j(e^{\omega_2x})^{n-k-j}}$$ $$\sum_{k=0}^n\sum_{j=0}^{n-k}\binom{n}{k,j,n-k-j}e^{\alpha x}$$ where $\alpha=\omega_0k+\omega_1j+\omega_2(n-k-j)$. Is there a way to simplify $\alpha$ anymore than this? And is this correct?
Note that $e^{\omega_0x}+e^{\omega_1x}+e^{\omega_2x}=1+\mathrm{e}^{2\pi i x/3}+\mathrm{e}^{2\cdot2\pi i x/3}$. Clearly $$ 1+\mathrm{e}^{2\pi i x/3}+\mathrm{e}^{2\cdot2\pi i x/3}=\frac{\mathrm{e}^{2\pi i x}-1}{\mathrm{e}^{2\pi i x/3}-1}=\mathrm{e}^{2\cdot2\pi i x/3}\cdot\frac{\mathrm{e}^{\pi i x}-\mathrm{e}^{-\pi i x}}{\mathrm{e}^{\pi i x/3}-\mathrm{e}^{-\pi i x/3}}=\mathrm{e}^{2\cdot2\pi i x/3}\cdot\frac{\sin (\pi x)}{\sin(\pi x/3)} \\=\mathrm{e}^{2\cdot2\pi i x/3}\cdot\frac{3\sin (\pi x/3)-4\sin^3(\pi x/3)}{\sin(\pi x/3)}=\mathrm{e}^{2\cdot2\pi i x/3}\big(3-4\sin^2(\pi x/3)\big). $$ Thus $$ (1+\mathrm{e}^{2\pi i x/3}+\mathrm{e}^{2\cdot2\pi i x/3})^n=\mathrm{e}^{4n\pi i x/3}\big(3-4\sin^2(\pi x/3)\big)^n=\mathrm{e}^{4n\pi i x/3}\left(\sum_{k=0}^n 3^k4^{n-k}\binom{n}{k}\sin^{2n-2k}(\pi x/3). \right) $$