Express as the product of four factors

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Homework Help Overview

The discussion revolves around the algebraic expression of the difference of squares, specifically focusing on expressing \(a^6 - b^6\) as the product of four factors.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various factorizations of the expression, questioning the requirement for four factors. Some suggest using the difference of squares and cubes, while others express confusion about the steps involved.

Discussion Status

Multiple interpretations of the problem are being explored, with some participants providing guidance on how to approach the factorization. There is a mix of understanding and confusion among participants regarding the correct method to achieve four factors.

Contextual Notes

Participants note that the original expression does not meet the requirement of having four factors, prompting further discussion on how to correctly factor it. There is also mention of potential grading implications based on the number of factors presented.

MorallyObtuse
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Homework Statement


Express as the product of four factors
Is this correct?
a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^3)
 
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MorallyObtuse said:

Homework Statement


Express as the product of four factors
Is this correct?
a^6 - b^6 = (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^3)
What you have written is true. It is not a "correct" response to the problem because, obviously, it does not have four factors. Start again. Think of a^6- b^6 as (a^3)^2- (a^3)^2.
 
I don't get it?!
 
Hi MorallyObtuse,
By the law of indices (a m)n=amxn
and so you can also represent it like this(like HallsofIvy said)
=(a^3)^2-(b^3)^2
And then you can simplyfy it like this
=(a^3-b^3)(a^3+b^3)
and finaly u have your 4 terms
(a-b) (a^2+ab+b^2) (a+b)(a^2-ab+b^2)
I think you get it.
 
1/2" said:
Hi MorallyObtuse,
By the law of indices (a m)n=amxn
and so you can also represent it like this(like HallsofIvy said)
=(a^3)^2-(b^3)^2
And then you can simplyfy it like this
=(a^3-b^3)(a^3+b^3)
and finaly u have your 4 terms
(a-b) (a^2+ab+b^2) (a+b)(a^2-ab+b^2)
I think you get it.

No I don't get it:biggrin::biggrin::biggrin:
 
a6 - b6 = (a3 + b3)(a3 - b3)
Each factor on the right can be further factored using known formulas for the sum of cubes and difference of cubes.
 
MorallyObtuse said:
I don't get it?!

MorallyObtuse said:
No I don't get it:biggrin::biggrin::biggrin:
Then you should talk to your teacher about it.
 
I did get it, just joking with 1/2" :biggrin:
 
And thanks very much :)

(a-b) (a^2+ab+b^2) (a+b)(a^2-ab+b^2)
 
  • #10
1 \times 1 \times 1 \times (a^6 - b^6)
 
  • #11
Heck, I can find 7 factors!
1 \times 1 \times 1 \times 1 \times 1 \times 1 \times (a^6 - b^6):biggrin:
 
  • #12
Yes, but that would be marked wrong because the problem asked for four factors. Borek's brilliant answer gets the gold star!
 
  • #13
I was hoping for extra credit because I went above and beyond the requirements.
 

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