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Finding the coefficients of a polynomial given some restriction

  • #1

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
b0cb62256bf80e1babd12ebc716ce9a9f6bb641c.png
be7630baf32627b0927dfb9d6ca49d3e409057a0.png
 

Attachments

Answers and Replies

  • #2
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There is another symmetry. Say ##z_1\in \mathbb{R}##, then ##z_{2,3}=x \pm iy##.
 
  • #3
33,286
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The images in post #1 are unreadable, at least by me.
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
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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.##
 

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