Mind your Ps and Qs

Just to make the algebra easier let

$p= 3P$ and $q=3Q$ so that the equation becomes

$z^3+3Pz^2 +3Qz +r=0$

Now make the substitution $z=x-P \dots(1)$ to get

$(x^3-3x^2 P+3x P^2-P^3)+3P(x^2-2Px+P^2)+3Q(x-P)+r=0$

Now collect like terms, combining all the constant terms into a single complex number, k.

$x^3 + (-3P+3P)x^2+(3P^2-6P^2+3Q)x+k=0$

Now if we set $Q=P^2 \dots(2)$

the equation becomes

$x^3=-k$

This equation has three roots placed symmetrically on a circle in the complex plane and so forming an equilateral triangle centred on the origin..

The transformation (1) shows that the z-solutions to the original problem are found by translating these x-solutions by the complex number -P. This simply slides the roots along the plane, maintaining them as an equilateral triangle!

To complete the problem we just need to translate the condition (2) back into the language of p and q, giving $\boxed{p^2=3q}$