For $a,b \in \mathbb{C}$, when is $a^3 = b^3$? As the title states. This seems as if it should be really simple, but I can't seem to figure it out.
Obviously, $a = c$ is a solution. I've tried something along the lines of 
$$(a+bi)^3 = (c+di)^3$$
but this hasn't really taken me anywhere. Am I missing something big here?
Apparently,
$$a = \frac{-1+\sqrt{3}\,i}{2}c$$ is also a solution, but for the life of me I can't seem to figure out how to get there myself.
Thanks in advance.
 A: You need $a^3 = b^3$ then $(\frac{a}{b})^3 = 1$. Let $z = \frac{a}{b}$ then $$z^3 = 1 \implies z^3-1=0 \implies (z-1)(z^2+z+1)=0 \implies z = 1,\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$$
Therefore
$$a = b ~~~\mbox{or}~~~a = \frac{-1+i\sqrt{3}}{2}b ~~~\mbox{or}~~~a = \frac{-1-i\sqrt{3}}{2}b $$
A: Polar form was made for this.
Let $a = r_a*e^{i\theta_a}$ and $b= r_b*e^{i\theta_b}$ so $a^3 = b^3$ means $r_a^3 e^{i3\theta_a} = r_b^3 e^{i3\theta_b}$ which means.
$r_a^3 = r_a^3; r_a, r_b \ge 0; r_a, r_b \in \mathbb R$.  So $r_a = r_b$.
And $ e^{i3\theta_a} =  e^{i3\theta_b}$ which means $3\theta_a = 3\theta_b \pm 2k\pi$.
So $\theta_a = \theta_b \pm \frac 23k\pi$
So $a^3 = b^3$ means $a = b*e^{i\frac 23k\pi}$
$e^{i\frac 23\pi} = $ a basic third-root of unity $= \cos (\arctan \frac 23 \pi) + i \sin (\arctan 23 \pi)= \frac {-1 + \sqrt{3}i}2 = \omega$ and $\omega^3 = 1$.
So the three solutions are $b = a; a*\omega; a*\omega^2$ where $\omega^2 = \frac {-1 - \sqrt{3}i}2$.
You can verify this be noting $(\frac {-1 + i\sqrt{3}}2)^3=1$
