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Question 52214This question is from textbook
: p.186
Factor:
(a+b)^6 - (a-b)^6
This question is from textbook
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! This is a good question! It is both the difference of two perfect squares and the difference of two perfect cubes. I am choosing to attack the squares first and then the cubes. There are three formulas that I am going to use.
u^2-v^2=(u+v)(u-v)
u^3-v^3=(u-v)(u^2+uv+v^2)
u^3+v^3=(u+v)(u^2-uv+v^2)
When I simplify, I'm going to need to foil or remember:
(u+v)^2=u^2+2uv+v^2
(u-v)^2=u^2-2uv+v^2
(u+v)(u-v)=u^2-v^2
So lets get started:
(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)
=((a+b)+(a-b))((a+b)^2-(a+b)(a-b)+(a-b)^2)((a+b)-(a-b))((a+b)^2+(a+b)(a-b)+(a-b)^2)
Now simplify and combine like terms:
=((a+b)+(a-b))((a^2+2ab+b^2)-(a^2-b^2)+(a^2-2ab+b^2))((a+b)-(a-b))((a^2+2ab+b^2)+(a^2-b^2)+(a^2-2ab+b^2))
=(a+b+a-b)(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2)(a+b-a+b)(a^2+2ab+b^2+a^2-b^2+a^2+2ab+b^2)
=(2a)(a^2+3b^2)(2b)(3a^2+b^2)
=(2a)(2b)(a^2+3b^2)(3a^2+b^2)
=4ab(a^2+3b^2)(3a^2+b^2)
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