I have an advanced algebra question... Factor (a + b)^6 - (a - b)^6. Please explain the methods, too!?

Question

Note: the ^ means "to the power of." I am particularly wondering whether I should factor a difference of squares or a difference of cubes to start.

Answer 1

Always start with difference of squares, because you can keep going with difference of squares; you usually have to stop with difference of cubes.

(a + b)^6 - (a - b)^6
= [((a + b)^3)^2 - ((a - b)^3)^2]
= [(a + b)^3 + (a - b)^3] * [(a + b)^3 - (a - b)^3]

Expand:
= [a^3 + 3a^2b + 3ab^2 + b^3 + a^3 - 3a^2b + 3ab^2 - b^3] *
[a^3 + 3a^2b + 3ab^2 + b^3 - a^3 + 3a^2b - 3ab^2 + b^3]

Combine terms:
= [2a^3 + 6ab^2] * [6a^2b + 2b^3]

Take out the common factors:
= 2a[a^2 + 3b^2] * 2b[3a^2 + b^2]
= 4ab(a^2 + 3b^2)(3a^2 + b^2)

And ya can't factor that any farther without using complex numbers, so I think that's it.

Answer 2

If you know something about binomial expansion and the binomial coefficient, or equivalently, Pascal's triangle, you can simplify the job.

You only need to keep the terms having b^k where k is odd

So, you'll end up with:

2{ 6C1a^5b^1 + 6C3a^3b^3 + 6C5a^1b^5 }

where nCk is the binomial coefficient or combinations.

Answer 3

Difference of squares is easier simply because the formula is shorter.

((a+b)^3 - (a-b)^3 ) ((a+b)^3 + (a-b)^3 )
=(6a^2b + 2b^3) ( 2a^3 + 6ab^2)
=(2b) (3a^2 + b^2) (2a) (a^2 + 3b^2)
=(4ab)(3a^2 + b^2) (a^2 + 3b^2)

Answer 4

do not do the difference of squares or cubes it will mess you up. (a+b)^2 is no the same as a^2+b^2.
just do (a+b)^2(a+b)(a+b)^3-(a-b)^2(a-b)^2(a-b)^2
(a^2+ab+b^2)(a+b)=a^3+2a^2b+2ab^2+b^3
the rest is too much for me to do now but you get the general idea

Answer 5

(a+b)^6-(a-b)^6 can beexpressed as a difference of squares as
[(a+b)^3]^2-[(a-b)^3]^2
=[(a+b)^3+(a-b)^3][(a+b)^3-(a-b)^3]
=[a^3+b^3+3a^2b+3ab^2+a^3-b^3-3a^2b+3ab^2]
[a^3+b^3+3a^2b+3ab^2-a^3+b^3+3a^2b-3ab^2]
=[2a^3+6ab^2][2b^3+6a^2b)
=[2a(a^2+3b^2][2b(b^2+3a^2]
=4ab(a^2+3b^2)(b^2+3a^2)