How Do You Factor Complex Algebraic Expressions Like This Olympiad Problem?

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SUMMARY

The discussion focuses on factoring the complex algebraic expression a(1-b²)(1-c²) + b(1-c²)(1-a²) + c(1-a²)(1-b²) - 4abc, which is presented as an Olympiad problem. Participants explore the use of algebraic identities and the cyclic symmetry of the equation to identify potential factors. The solution involves recognizing independent terms in the expanded form and suggests that factors can be derived from the expression (abc - a - b - c). Additionally, resources for further reading on factorization techniques are provided.

PREREQUISITES
  • Understanding of algebraic identities
  • Familiarity with polynomial expansion
  • Knowledge of cyclic symmetry in algebra
  • Experience with mathematical Olympiad problems
NEXT STEPS
  • Study advanced factorization techniques in algebra
  • Learn about cyclic and symmetric polynomials
  • Practice solving Olympiad-style algebra problems
  • Explore resources on polynomial identities and their applications
USEFUL FOR

Mathematics students, educators, and enthusiasts interested in advanced algebra, particularly those preparing for mathematical Olympiads or seeking to enhance their problem-solving skills in algebraic expressions.

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Homework Statement


Factorate the expression a(1-b²)(1-c²) + b(1-c²)(1-a²) + c(1-a²)(1-b²) - 4abc

Homework Equations


Algebric identities

The Attempt at a Solution


It seems to be an olympic problem, and I can't find the right factors for it.
I would really appreciate the help and, if possible, could you indicate me some articles about factorizations, like this one, for mathematic olympiads.
Thanks
 
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This was a bit hard to crack. I'll show you my thought chain. By the nature of the equation you see there are going to be independent a, b, c terms in the equation when expanded. This hints to a final factor containing constant term (±1) multiplied by another which contains independent terms (± a ± b ± c). After that I couldn't make any conclusions to decide the factors, and hence expanded the equation, which gives,

a²b²c + b²c²a + c²a²b - a²b - a²c - b²a - b²c - c²a - c²b + a + b + c - 4abc

Taking out ab, bc and ca as factors, and by cyclic symmetry in the equation,

ab(abc - a - b) + bc(abc - b - c) + ca(abc - c - a) + (a + b + c - 4abc)

Now if the factors (abc - a - b) , (abc - b - c) and (abc - c - a) could all be made of the form (abc - a - b - c) you can factor it out from all the terms. See if you can do this.

I'm not aware of any specific articles on factorization. Just play around with enough problems to give you an hint of what you can do in various situations.

PS: I'm curious to see if there are more elegant solutions to this from someone, than expanding out the whole equation.

Edit: Google gave these articles,
http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx
http://www.qc.edu.hk/math/Resource/AL/Cyclic%20and%20symmetric%20polynomials.pdf
 
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