This is exercise $1$ on page $45$ of Tristan Needham's Visual Complex Analysis:
1. The roots of a general cubic equation in $X$ may be viewed (in the $XY$- plane) as the intersection of thhe $X$- axis with the graph of a cubic of the form,
$$Y = X^3 + AX^2 + BX + C$$
$(i)$ Show that the point of inflection of the graph occurs at $X = -\dfrac A3$
$(ii)$ Deduce (geometrically) that the substitution $X = (x-\dfrac A3)$ will reduce the above equation to the form $Y = x^3 + bx + c$.
$(i)$ is easy enough by using the second derivative. (is there another way though?)
$(ii)$ I know that making said substitution will center the graph so that the inflection point will lie on the $Y$ axis. However, I'm not sure what to do beyond that. By staring at the graph below, I realized just now that it would be reasonable visually to argue $Y'(x) = Y'(-x)$, so the derivative has to be a parabola of the form $x^2 + \gamma$, which means that $Y$ is a depressed cubic. However, I'm not sure if this is a geometric enough approach.
I also tried to view of $0 = X^3 + AX^2 + BX + C$ as $x^3$ intersecting a parabola, and of $0 = x^3 + bx+c$ as $x^3$ intersecting a line; but I'm not sure how helpful this is because then the substitution $X = x - \dfrac A3$ loses its easy visual meaning.
Cubic:
Depressed Cubic:
