Which method is easier and why?

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Discussion Overview

The discussion revolves around the problem of factoring the expression x^6 - y^6 using two different methods: as a difference of squares and as a difference of cubes. Participants explore which method may be easier and seek guidance on how to begin the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants suggest that applying the difference of squares leads to simpler next steps, providing the factorization as (x^3+y^3)(x^3-y^3).
  • Others present the complete factorization resulting from the difference of squares method, detailing the steps and resulting factors.
  • There is an alternative approach mentioned using the difference of cubes, leading to a different factorization, but it is noted that this method may not be as straightforward.
  • Participants express that both methods can be applied, but the ease of subsequent steps may vary depending on the chosen method.

Areas of Agreement / Disagreement

Participants generally agree that both methods can be applied, but there is no consensus on which method is definitively easier, as opinions vary based on individual perspectives on the complexity of the steps involved.

Contextual Notes

The discussion includes various mathematical steps and assumptions that may not be fully resolved or universally accepted, particularly regarding the ease of each factoring method.

mathdad
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I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?
 
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Excellent!
 
RTCNTC said:
I found the following problem in Section 1.3 of my Precalculus textbook by David Cohen.

Factor x^6 - y^6 as a difference of squares and then as a difference of cubes.

1. Which method is easier and why?

2. Can someone get me started?

you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable
 
kaliprasad said:
you can apply either way. but next steps become simpler if you apply difference of square

$x^6-y^6= (x^3+y^3)(x^3-y^3)$ 1st term is sum of cubes and second is difference of cubes and you get

$x^6-y^6= (x^3+y^3)(x^3-y^3)= (x+y)(x^2-xy+y^2)(x-y)(x^3+xy+y^2)$

had you chosen difference of cubes you would have got

$x^6-y^6 = (x^2-y^2)(x^4+x^2y^2+y^4)$

factoring $x^4+x^2y^2+y^4 = (x^4+2x^2y^2+y^4)- x^2y^2 = (x^2+y^2)^2 - (xy)^2 = (x^2+y^2+xy)(x^2+y^2-xy)$

above may not be straight forward but doable

Nice work in every sense.
 

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