Problem source

Show that  the distinct complex numbers $\alpha$, $\beta$ and $\gamma$ represent the vertices of an equilateral triangle (in clockwise or anti-clockwise order) if and only if
\[
\alpha^2 + \beta^2 +\gamma^2 -\beta\gamma - \gamma \alpha -\alpha\beta =0\,.
\]
Show that the roots of the equation 
\begin{equation*}
z^3 +az^2 +bz +c=0
\tag{$*$}
\end{equation*}
represent the vertices of an equilateral triangle if and only if $a^2=3b$. 
Under the transformation  $z=pw+q$, where $p$ and $q$ are given complex numbers with $p\ne0$, the equation ($*$) becomes
\[
w^3 +Aw^2 +Bw +C=0\,.
\tag{$**$}
\]
Show that if the roots of equation $(*)$ represent the vertices of an equilateral triangle, then  the roots of equation $(**)$ also represent  the vertices of an equilateral triangle.

Solution source

The complex numbers represent an equilateral triangle iff $\gamma$ is a $\pm 60^\circ$ rotation of $\beta$ around $\alpha$, ie 

\begin{align*}
&& \gamma - \alpha &= \omega(\beta - \alpha) \\
\Leftrightarrow && \omega &= \frac{\gamma - \alpha}{\beta - \alpha} \\
\Leftrightarrow && -1 &= \left (\frac{\gamma - \alpha}{\beta - \alpha} \right)^3 \\
\Leftrightarrow && -(\beta - \alpha)^3 &=(\gamma - \alpha)^3 \\
\Leftrightarrow && 0 &= (\gamma-\alpha)^3+(\beta-\alpha)^3 \\
&&&= \gamma^3-3\gamma^2\alpha +3\gamma\alpha^2-\alpha^3 +\beta^3-3\beta^2\alpha+3\beta\alpha^2-\alpha^3 \\
&&&= (\beta + \gamma - 2\alpha)(\alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta) \\
\Leftrightarrow && 0 &= \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta
\end{align*}

The roots of the equation $z^3+az^2+bz+c = 0$ represents the vertices of an equilateral triangle iff $a^2-3b = (\alpha+\beta+\gamma^2) - 3(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 - \alpha\beta - \beta\gamma - \gamma \delta = 0$ as erquired.

Suppose $a^2 = 3b$, then consider $z = pw +q$, we must have

\begin{align*}
&& 0 &= (pw+q)^3+a(pw+q)^2 + b(pw+q)+c \\
&&&= p^3w^3 +(3p^2q+ap^2)w^2+(3pq^2+2apq+bp)w+(q^3+aq^2+bq+c) \\
&&&= p^3w^3+p^2(3q+a)w^2+p(3q^2+2aq+b)w+(q^3+aq^2+bq+c) \\
\end{align*}

We need to check if $\left(\frac{3q+a}{p} \right)^2 = 3 \left (\frac{3q^2+2qa+b}{p^2} \right)$. Clearly the denominators match, so consider the numerators

\begin{align*}
&& (3q+a)^2 &= 9q^2+6aq+a^2 \\
&&&= 9q^2+6aq+3b \\
&&&= 3(3q^2+2qa+b)
\end{align*}

as required