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

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?

• Hint : 3 complex numbers $z_1,z_2,z_3$ form an equilateral triangle if and only if $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$. Commented Nov 6, 2022 at 3:04
• I have taken the liberty to modify the title of your question in order for it to reflect the content of the question. Do you agree ? Commented Nov 6, 2022 at 17:50
• Jean, that's OK. Commented Nov 7, 2022 at 6:35

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.

• Once more an anonymous downvoter for a neat answer (so I think...). What a plague on this site ! Downvoting shouldn't be anonymous. Commented Nov 7, 2022 at 10:59