For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane iff $\Re(w)=-\frac12$. For $w\in\mathbb{C}$, prove that $w^3$ lies on the line passing through 1 and $w$ in the complex plane if, and only if, $\Re(w)=-\frac12$.
I came across this fact while messing about with spirals of powers of a complex number. I've tried parametrising the straight line with a real number $t$, so, for example, $z(t)=(1-t)+tw$. But alas to no avail! Any insights are appreciated.
Here is a link to a Desmos graph showcasing the phenomenon:
https://www.desmos.com/calculator/gsugcdlkc9

A different proof:
Let $\omega = x + iy\in\mathbb{C}, y \neq 0, \omega \neq 1$
$\omega^3 = (x^3 - 3xy^2) + i(3x^2y - y^3)$
The three points $\omega \equiv(x, y)$, $(1,0)$ and $\omega^3 \equiv (x^3 - 3xy^2, 3x^2y - y^3)$ are collinear, iff the following determinant is $0$
\begin{vmatrix}
x & y & 1 \\ 
1 & 0 & 1 \\ 
x^3 - 3xy^2 & 3x^2y - y^3 & 1 
\end{vmatrix} 
Simplifying,
$y(2x+1)[(x-1)^2 + y^2)] = 0$
Since $y \neq 0$, this is possible iff $ x = -\frac{1}{2}$
A shorter proof:
Let $\omega=x + iy, y \neq 0, \omega \neq 1$
$1,\omega,\omega^3$ are collinear iff $\frac{\omega^3-\omega}{\omega-1} = \omega^2 + \omega + 1$ is real.
Since $\omega^2 + \omega + 1 = (x^2 + x - y^2 + 1) + i y(2x + 1)$, we must have $y(2x + 1) = 0$
Now $y \neq 0 \implies x = -\frac{1}{2}$ 
