Express as the product of four factors

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The discussion revolves around factoring the expression a^6 - b^6 into four factors. Initially, a user presents the factorization as (a - b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5), which is incorrect since it only has two factors. Participants suggest using the difference of squares and the sum and difference of cubes to arrive at the correct factorization: (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2). The conversation highlights the importance of adhering to the requirement of expressing the expression as a product of exactly four factors. Ultimately, the correct approach is confirmed, emphasizing the need for clarity in mathematical problem-solving.
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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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