Finding the coefficients of a polynomial given some restriction

[MatejNeumann]

Homework Statement

Find all $a,b,c\in\mathbb{R}$ for which the zeros of the polynomial $az^3+z^2+bz+c=0$ are in this relation $$z_1^3+z_2^3+z_3^3=3z_1z_2z_3$$

Homework Equations

we know that if we have a polynomial of degree 3 the zeroes have relation in this case
$z_1+z_2+z_3=-1/a$
$z_1z_2+z_1z_3+z_2z_3=b/a$
$z_1z_2z_3=-c/a$

The Attempt at a Solution

I've tried doing something with the vietto rules but I have not really gotten anything. Any tip on how to even start the problem would be really appreciated

 

[fresh_42]

There is another symmetry. Say $z_1\in \mathbb{R}$, then $z_{2,3}=x \pm iy$.

 

[Mark44]

The images in post #1 are unreadable, at least by me.

 

[Ray Vickson]

Homework Statement

Find all $a,b,c\in\mathbb{R}$ for which the zeros of the polynomial $az^3+z^2+bz+c=0$ are in this relation $$z_1^3+z_2^3+z_3^3=3z_1z_2z_3$$

Homework Equations

we know that if we have a polynomial of degree 3 the zeroes have relation in this case
$z_1+z_2+z_3=-1/a$
$z_1z_2+z_1z_3+z_2z_3=b/a$
$z_1z_2z_3=-c/a$

The Attempt at a Solution

I've tried doing something with the vietto rules but I have not really gotten anything. Any tip on how to even start the problem would be really appreciated
View attachment 232485 View attachment 232486

If $z_i$ is a root then $$z_i^3 = -\frac{c}{a} - \frac{b}{a} z_i - \frac{1}{a} z_i^2,$$
and we can find $\sum z_i^2$ from the expansion of $(z_1+z_2+z_3)^2.$