X^6 - y^6 As Difference of Cubes

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SUMMARY

The expression x^6 - y^6 can be factored as a difference of cubes using the formula (x^3 - y^3)(x^3 + y^3). Further breakdown reveals that x^3 - y^3 factors into (x - y)(x^2 + xy + y^2) and x^3 + y^3 factors into (x + y)(x^2 - xy + y^2). An alternative approach involves recognizing x^6 - y^6 as (x^2)^3 - (y^2)^3, leading to the factorization (x^2 - y^2)((x^2)^2 + x^2y^2 + (y^2)^2), which simplifies to (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2).

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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?
 
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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
 
Great. Your previous reply is more than enough in terms of this question.
 
RTCNTC said:
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).
 
Thank you everyone.
 

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