Solving a system of 3 nonlinear equations

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    Nonlinear System
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Discussion Overview

The discussion revolves around solving a system of three nonlinear equations involving variables x, y, and z, with specific relationships defined by the equations a = xyz, b = xy + xz + yz, and c = x + y + z. The focus includes exploring methods for finding solutions, particularly through the manipulation of these equations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests expressing x in terms of c, y, and z, leading to a quadratic equation for y or z, and hopes for a "nice" solution.
  • Another participant counters that the equation is not quadratic and claims it leads to a cubic equation z^3 - cz^2 + bz - a = 0, indicating a different approach to finding z as a function of a, b, and c.
  • A further reply asserts that the cubic equation appears "nice" and implies that finding its roots could facilitate determining x and y in terms of a, b, and c.
  • There is a mention of uncertainty regarding the method for finding the roots of a cubic function, indicating a gap in knowledge among participants.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the equations involved, with some asserting the presence of a quadratic equation while others maintain that it is a cubic equation. The discussion remains unresolved regarding the best approach to solve the system.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the equations and the methods for solving them, particularly concerning the transition from quadratic to cubic forms and the techniques for finding roots of cubic equations.

EM_Guy
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a = xyz
b = xy+xz+yz
c = x + y + z

How do you solve x, y, and z?
 
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x=c-y-z
b=(c-y-z)y + (c-y-z)z + yz = c(y+z)-y^2-zy-z^2
Solve this quadratic equation for y (or z), use both in a=xyz and hope that it has a nice solution?
 
It is not a quadratic equation. And it is not a "nice" solution.

I have determined that z^3-cz^2+bz-a = 0. So, if we can find the roots of the cubic function, then we have z as a function of a, b, and c. Then, it should be straightforward to find x and y in terms of a, b, and c.

But I forget how to find the roots of a cubic function.
 

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