Express as the product of four factors

[MorallyObtuse]

Homework Statement

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

 

[HallsofIvy]

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 $(a^3)^2- (a^3)^2$.

 

[MorallyObtuse]

I don't get it?!

 

[1/2"]

Hi MorallyObtuse,
By the law of indices (a ^m)ⁿ=a^mxn
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.

 

[MorallyObtuse]

Hi MorallyObtuse,
By the law of indices (a ^m)ⁿ=a^mxn
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

 

[Mark44]

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.

 

[HallsofIvy]

I don't get it?!

No I don't get it

Then you should talk to your teacher about it.

 

[MorallyObtuse]

I did get it, just joking with 1/2"

 

[MorallyObtuse]

And thanks very much :)

$(a-b) (a^2+ab+b^2) (a+b)(a^2-ab+b^2)$

 

[Borek]

$$1 \times 1 \times 1 \times (a^6 - b^6)$$

 

[Mark44]

Heck, I can find 7 factors!
$$1 \times 1 \times 1 \times 1 \times 1 \times 1 \times (a^6 - b^6)$$

 

[HallsofIvy]

Yes, but that would be marked wrong because the problem asked for four factors. Borek's brilliant answer gets the gold star!

 

[Mark44]

I was hoping for extra credit because I went above and beyond the requirements.