$$\left ( 1 + \cos \frac{π}{6} + i \sin\frac{π}{6} \right )^6$$
For a question like this, the first thing I would think of using is the DeMoivre's Theorem, however, with the entire real part of it being (1+cos(π/6)), I'm not sure if the rule will apply.
How do I approach this?
Observe that $\cos\frac\pi6+i\sin\frac\pi6$ is the unit vector of argument $\pi/6$. It is easy to see that this vector when added to 1 gives a vector of argument $\pi/12$ and length $\sqrt{1^2+1^2-2\cos\frac{5\pi}6}=\sqrt{2+\sqrt3}$.
Taking this to the sixth power multiplies the argument by six (hence it becomes $\frac\pi2$) and raises the length to the sixth power (hence $(2+\sqrt3)^3=8+3\cdot4\sqrt3+3\cdot6+3\sqrt3=26+15\sqrt3$). The final result is $(26+15\sqrt3)e^{i\pi/2}$ or $(26+15\sqrt3)i$.
Hint : remark that, for any $\theta \in \mathbb{R}$, we have $$1+e^{i \theta}= e^{i 0}+e^{i \theta}=e^{i \theta/2}(e^{-i \theta/2 }+e^{i \theta/2})=2e^{i \theta/2} cos(\theta/2).$$
Hint:
To have lighter notations, set $\;u=\mathrm e^{\tfrac{i\pi}6}$, $\bar u=\mathrm e^{-\tfrac{i\pi}6}$, and rewrite the sum as \begin{align} (1&+u)^6+(1+\bar u)^6= \\ 2&+6(u+\bar u)+15(u^2+\bar u^2)+20(u^3+\bar u^3)++15(u^4+\bar u^4)+6(u^5+\bar u^5)+u^6+\bar u^6. \end{align} You can compute the sums of powers gradually. Note that $u+\bar u=\sqrt 3$, that $u^3=i$, $\bar u^3=-i$, so $u^3+\bar u^3=0$, and that $u\,\bar u=1$.
Edit:
We can use the same method (expanding the binomial) to the new formulation of the question: $$(1+u)^6=1+6u+15u^2+20u^3+15u^4+6u^5+u^6.$$ Note in addition to the above that:
so $\;(1+u)^6=1+(26+15\sqrt 3)i.$
