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. 
 A: You have the equations 
$$a^3-3ab^2=1\\3a^2b-b^3=0$$
Having identified $a=1,b=0$ as one solution, you can now note that $b\neq 0$ in any other solutions.  You can then divide the second one by $b$ and get 
$$3a^2=b^2$$
and plug that into the first to get
$$a^3-9a^3=1\\a=-\frac 12\\b^2=\frac 34\\b=\pm \frac{\sqrt 3}2$$
A: If $a+bi$ is a solution and $b=0$, then your equations are simply$$\left\{\begin{array}{l}a^3=1\\0=0,\end{array}\right.$$whose only solution is $(a,b)=(1,0)$. In other words, $a+bi=1$.
Now, suppose that $b\neq0$. Then your system is equivalent to$$\left\{\begin{array}{l}a^3-3ab^2=1\\3\left(\frac ab\right)^2=1,\end{array}\right.$$But then $\frac ab=\pm\frac1{\sqrt 3}$, which is equivalent to $a=\pm\frac b{\sqrt 3}$. Now, replace $a$ with $\pm\frac{b}{\sqrt 3}$ in the first equation to solve the system.
A: We have 
\begin{eqnarray*}
a^3-3ab^2=1 \\
3a^2b-b^3=0.
\end{eqnarray*}
From the second equation we have $b=0$ (giving the obvious solution) or $3a^2=b^2$, sub this into the first equation and we get $a=-1/2$ and then sub this back into $3a^2=b^2$ and we have the other two roots of unity
\begin{eqnarray*}
-\frac{1}{2} \pm \frac{ \sqrt{3}}{2}.
\end{eqnarray*}
A: Do you know De Moivre's law?
For any complex number $z$ we can write it in polar.
$z = |z|(\cos\theta + i\sin \theta)$
And from here
$z^n = |z|^n (\cos n\theta + i\sin n\theta)$
In this problem
$z^3 = |z|^3 (\cos 3\theta + i\sin 3\theta) = 1 + 0i$
For what theta does
$\cos 3\theta = 1$ and $\sin 3\theta = 0$
$3\theta = 0, 2\pi, 4\pi, \cdots\\
\theta = \frac {2\pi}{3}, \frac {4\pi}{3}, 2\pi$
$z = -\frac {\sqrt {3}}{2} + \frac {i}{2}, -\frac{\sqrt {3}}{2} - \frac {i}{2}, 1$
A: 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 $(1,0), (-\frac12,\frac{\sqrt3}2), (-\frac12,-\frac{\sqrt3}2)$.
