X^6 - y^6 As Difference of Cubes

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Discussion Overview

The discussion revolves around factoring the expression x^6 - y^6 as a difference of cubes. Participants explore various approaches to factor this expression, including the application of the difference and sum of cubes formulas.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes the initial factorization of x^6 - y^6 as (x^3 - y^3)(x^3 + y^3) and seeks confirmation on proceeding with further factorization.
  • Another participant suggests that the intention is to use the difference of cubes formula, referencing a previous discussion on the topic.
  • A later reply reiterates the factorization and expands it by applying the difference and sum of cubes, providing the complete factorization as (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2).
  • Additionally, an alternative approach is mentioned, where x^6 - y^6 is factored as (x^2)^3 - (y^2)^3, leading to a different but equivalent factorization involving (x^2 - y^2) and further factors.

Areas of Agreement / Disagreement

Participants appear to agree on the validity of the factorization methods presented, but there are multiple approaches discussed without a clear consensus on a preferred method.

Contextual Notes

The discussion includes various assumptions about the application of algebraic identities and the conditions under which these factorizations hold. Some steps in the reasoning may depend on the definitions and properties of the expressions involved.

mathdad
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Factor x^6 - y^6 as a difference of cubes.

Solution:

(x^3 - y^3)(x^3 + y^3)

I now proceed by applying the difference and sum of cubes accordingly, right?
 
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I'm thinking the intention here is to first use the difference of cubes formula...however, I treated that case in your other thread regarding this expression. :D
 
Great. Your previous reply is more than enough in terms of this question.
 
RTCNTC said:
Factor x^6 - y^6 as a difference of cubes.

Solution:

(x^3 - y^3)(x^3 + y^3)

I now proceed by applying the difference and sum of cubes accordingly, right?
Yes, of course, you can do that. Further, x^3- y^3= (x- y)(x^2 + xy+ y^2) and x^3+ y^3= (x+ y)(x^2- xy+ y^2) so that can be written as (x- y)(x+ y)(x^2+ xy+ y^2)(x^2- xy+ y^2).

As you saw in the other thread, you can also do it "the other way around". x^6- y^6= (x^2)^3- (y^2)^3= (x^2- y^2)((x^2)^2+ x^2y^2+ (y^2)^2)= (x^2- y^2)(x^4+ x^2y^2+ y^2)= (x- y)(x+ y)(x^2+ xy+ y^2)(x^2- xy+ y^2).
 
Thank you everyone.
 

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