Prove the equality (Taylor series).

Hint: (A followup to Lord Shark the Unknown's observation) You're already half-way there. You've established that \begin{align} \frac{1}{3}\left(e^x + 2e^\frac{-x}{2}\cos \frac{x\sqrt{3}}{2}\right) &=\frac{1}{3}\left(e^x + e^\frac{x\left(-1+i\sqrt{3}\right)}{2}+ e^\frac{x\left(-1-i\sqrt{3}\right)}{2}\right)\\ &= \frac{1}{3}\left(e^{z_1 x} + e^{z_2x} + e^{z_3}x \right)\ , \end{align} where $\ z_1=1\ $, $\ z_2=\frac{x\left(-1+i\sqrt{3}\right)}{2}\ $, and $\ z_3=\frac{\left(-1-i\sqrt{3}\right)}{2}\ $ are the three cube roots of unity. If you now use the expansions \begin{align} e^{z_ix}&= \sum_{n=0}^\infty\frac{z_i^nx^n}{n!}\\ &=\sum_{n=0}^\infty\left(\frac{z_i^{3n}x^{3n}}{(3n)!}+\frac{z_i^{3n+1}x^{3n+1}}{(3n+1)!}+\frac{z_i^{3n+2}x^{3n+2}}{(3n+2)!}\right)\\ &=\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}\left(1 +\frac{z_ix}{3n+1}+\frac{z_i^2x^2}{3n+2}\right)\ , \end{align} and the observations that $\ z_1^2=z_2\ $, $\ z_2^2=z_1\ $, and $\ z_1 + z_2 + z_3=0\ $, you should be able to complete the demonstration.