Prove that $\left(\frac {-1+\sqrt {-3}}{2}\right)^n + \left(\frac {-1-\sqrt {-3}}{2}\right)^n$ follows this pattern 
Prove that:
  $$\left(\dfrac {-1+\sqrt {-3}}{2}\right)^n + \left(\dfrac {-1-\sqrt {-3}}{2}\right)^n=\begin{cases}
2, & \textrm { if } n \textrm { is a multiple of 3},\\
-1, & \textrm { if } n \textrm { is any other integer}
\end{cases}$$

My Attempt:
$$\dfrac {-1+\sqrt {-3}}{2}=\dfrac {-1+i\sqrt {3}}{2}$$ which is a complex cube root of unity. 
Let $\omega = \dfrac {-1+i\sqrt {3}}{2}$.
Similarly, $\omega^2=\dfrac {-1-i\sqrt {3}}{2}$

Alternative solution: the two complex numbers $\frac{-1\pm\sqrt{3}i}{2}$ are the roots of $x^2+x+1=0.$ Consider a recursive formula
$$a_{n+2}+a_{n+1}+a_n=0,$$
with $a_0=2$ and $a_1=-1$ agreeing with $\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$ at $n=0$ and $1$. Then from the characteristic root method, $a_n=\left(\frac{-1+\sqrt{3}i}{2}\right)^n+\left(\frac{-1-\sqrt{3}i}{2}\right)^n$. We can then prove the periodic-$3$ pattern using mathematical induction. 
