Collinearity of the three roots of a cubic equation in the complex plane The three roots of the equation $x^3+bx+c=0\;(b,c\in\Bbb C,c\ne0)$ in the complex plane are collinear iff$$k\in\Bbb R\land k\le-\frac{27}4$$where $k=\frac{b^3}{c^2}$.
Original post on Math Entertainment BBS (The original question was to find a condition for a root to be the midpoint of the other two roots, and was solved using Vieta's theorem, then OP extended the question to collinearity)
 A: Given three distinct points $\,r_1,r_2,r_3\in\Bbb C,\,$ define the ratio
$\,\lambda:=\frac{-r_2}{r_1}.\,$ If the three
points are the roots of a cubic equation $\,x^3+bx+c=0,\,c\ne 0\,$ then
$$ r_1 \ne 0, \;\; r_2 = -\lambda r_1,\;\;
   r_3 = -(1-\lambda)r_1, \\
 b = -r_1^2(1-\lambda+\lambda^2), \;\;
 c = -r_1^3\lambda(1-\lambda),\\ 
 k :=  \frac{b^3}{c^2} = \frac{-(1-\lambda+\lambda^2)^3}
   {(\lambda(1-\lambda))^2} = \frac{-j(\tau)}{256} =
\frac{-27}{4} J(\tau), \\
 k\!+\!\frac{27}4 \!=\! \frac{27}4 (1\!-\!J(\tau))
 \!=\! -\left(\frac{(1\!+\!\lambda)(1\!-\!\lambda/2)
 (1\!-\!2\lambda)}{\lambda(1-\lambda)}\right)^2
 \!=:\! -L(\lambda)^2$$
where $\,j(\tau)=1728J(\tau)\,$ is the
$j$-invariant
related to the
modular lambda $\,\lambda(\tau).\,$
Note that the three points depend on $\,\lambda\,$ as
$$ (r_1,r_2,r_3) \sim (1,-\lambda,-1+\lambda). $$
The three points are collinear iff $\,\lambda\,$ is real.
Thus if $\,\lambda\,$ is real, then $\,L(\lambda)\,$ is real
which implies that $\,k+\frac{27}4 \le 0\,$ and
$\,k\,$ is real.
In the converse case, assume that $\,k\,$ is real and
$\,k\le -\frac{27}4.\,$ Then $\,J(\tau)\ge 1.\,$
Use knowledge of the mapping properties of $\,j(\tau)\,$ in
the fundamental region to imply that $\,\tau = i\,y\,$ where
$\,y\ge 1.\,$ The corresponding value of
$\,L(\lambda(\tau))\ge 0\,$ implies that
$\,k+\frac{27}4 \le 0\,$ and $\,k\,$ is real. This part of
the proof uses mapping properties of $\,j(\tau)\,$ but there
may be a proof using just the mapping properties of the
rational function $\,L(\lambda).\,$
