$a^2=3b$ is a necessary and sufficient condition for the roots of $x^3+ax^2+bx+c=0$ to constitute an equilateral triangle.

Question

The three roots of the equation $x^{3}+ax^{2}+bx+c=0$ (where $a, b, c$ are given complex numbers) are represented on the Argand diagram by the points $A, B, C$. Prove that $ABC$ is an equilateral triangle if and only if $a^{2}=3b$. Could someone help me to prove the converse part?

Let $z_1, z_2, z_3$ be the roots of the given equation. $z_1 + z_2 + z_3 = -a$ and $z_1z_2 + z_2z_3 + z_3z_1=b$. Using interior angle of an equilateral triangle is $\frac {\pi}{3}$. $(z_1-z_3)(z_1-z_2)=(z_2-z_3)(z_3-z_2)$ gives $a^{2}=3b$.

May I have some sort of working out to prove the converse of the given result?

Answer 1

Let us assume that $a^2=3b$. Replacing $b$ by

$$b=\frac{a^2}{3}$$

into the third degree equation, it becomes:

$$\left(x+\frac{a}{3}\right)^3=-c \tag{1}$$

Let $d$ be a cubic root of $-c$ (assumed nonzero). Then (1) can be written under the form:

$$\left(\underbrace{\frac{x+\frac{a}{3}}{d}}_Z\right)^3=1\tag{2}$$

The solutions to equation (2)

• in terms of variable $Z$, are the three third roots of unity, giving rise to the "canonical" equilateral triangle $1,e^{2i \pi/3}, e^{4i \pi/3}$.

• in terms of variable $x$, are images of the previous ones by transformation $x=dZ-\frac{a}{3}$ which preserves in particular angles, giving as well an equilateral triangle.