<s>[INST] Finding the third roots of unity by equating $(a + bi)^3$ to $1$ I know what the third roots of unity are but I want to solve this exercise:

Complex numbers can be written as $a + bi$. Simplify $(a + bi)^3$ and
  equate the real part to $1$ and the imaginary part to $0$ to find the
  three roots of unity.

So I expanded $(a + bi)^3$ to get $(a^3 - 3ab^2) + (3a^2b - b^3)i$. 
I know that $a = 1$ and $b = 0$ is a solution just by looking at it. Now is there some sort of clever way to get another root of unity? I tried to do it with just algebra but it ended up being very messy. I was wondering whether there is some neat solution to this. 
 [/INST] Factoring a difference of cubes, we have
\begin{align}
(a+bi)^3&=1\\
(a+bi)^3-1&=0\\
(a+bi-1)((a+bi)^2+(a+bi)+1)&=0\\
(a-1+bi)(a^2+2abi-b^2+a+bi+1)&=0.
\end{align}
So, we have either $$\operatorname{Re}(a-1+bi)=1\ \wedge\ \operatorname{Im}(a-1+bi)=0 \implies (a,b)=(1,0)$$
or
$$\operatorname{Re}(a^2+2abi-b^2+a+bi+1)=1\ \wedge\ \operatorname{Im}(a^2+2abi-b^2+a+bi+1)=0,$$
giving us the system of equations
\begin{align}
a^2-b^2+a+1&=1\tag1\\
2ab+b&=0.\tag2\end{align}
From $(2)$, we have $b(2a+1)=0\implies a=-\frac12$.  Substituting this into $(1)$, we have
$$\left(-\frac12\right)^2-b^2+\left(-\frac12\right)=0\implies b=\pm\frac{\sqrt3}2.$$
Therefore, the three roots of unity are $(1,0), (-\frac12,\frac{\sqrt3}2), (-\frac12,-\frac{\sqrt3}2)$.
</s>