jmc

Expanding both sides gives $$z^3-(r+s+t)z^2+(rs+st+rt)z-rst=z^3-c(r+s+t)z^2+c^2(rs+st+rt)z-c^{3}rst.$$Since this equation holds for all $z,$ we must have $$\{\begin{array}{rl}-(r+s+t)& =-c(r+s+t),\\ rs+st+rt& =c^2(rs+st+rt),\\ -rst& =-c^{3}rst.\end{array}$$If none of $c,c^2,c^3$ are equal to $1,$ then these equations imply that $$r+s+t=rs+st+rt=rst=0.$$Then $r,s,t$ are the roots of the polynomial $z^3-0z^2-0z-0=z^3,$ so $r=s=t=0,$ which contradicts the fact that $r,s,t$ must be distinct. Therefore, at least one of the numbers $c,c^2,c^3$ must be equal to $1.$

If $c=1,$ then all three equations are satisfied for any values of $r,s,t.$ If $c^2=1,$ then the equations are satisfied when $(r,s,t)=(0,1,-1).$ If $c^3=1,$ then the equations are satisfied when $(r,s,t)=\left(1,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right).$ Therefore, all such $c$ work. The equations $c=1,$ $c^2=1,$ and $c^3=1$ have a total of $1+2+3=6$ roots, but since $c=1$ satisfies all three of them, it is counted three times, so the number of possible values of $c$ is $6-2=\boxed{4}.$