MHB X^6 - y^6 As Difference of Cubes

[mathdad]

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?

 

[MarkFL]

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

 

[mathdad]

Great. Your previous reply is more than enough in terms of this question.

 

[HallsofIvy]

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).

 

[mathdad]

Thank you everyone.