F.3 factorization *hurry
回答 (2)
2008-01-04 6:28 am
✔ 最佳答案
(a+b)^3 - 27(a-b)^3= (a+b)^3 - [3(a-b)]^3
By the identity a^3 - b^3 = (a-b)(a^2 + ab + b^2)
(a+b)^3 - [3(a-b)]^3
= [(a+b) - 3(a-b)] {(a+b)^2 + (a+b)[3(a-b)] + [3(a-b)]^2}
= (a+b-3a+3b)(a^2 + 2ab + b^2 + 3a^2 - 3b^2 + 9a^2 - 18ab + 9b^2)
= (4b-2a)(13a^2 - 16ab + 7b^2)
= 2(2b-a)(13a^2 - 16ab + 7b^2)//
2008-01-04 6:22 am
(a+b)^3-27(a-b)^3
=(a+b)^3-[3(a-b)]^3
=[(a+b)-3(a-b)]{(a+b)^2+(a+b)(3)(a-b)+[3(a-b)]^2}
=(a+b-3a+3b)[a^2+2ab+b^2+3a^2-3b^2+9(a^2-2ab+b^2)]
=(4b-2a)(13a^2-16ab+7b^2)
=2(2b-a)(13a^2-16ab+7b^2)
=(a+b)^3-[3(a-b)]^3
=[(a+b)-3(a-b)]{(a+b)^2+(a+b)(3)(a-b)+[3(a-b)]^2}
=(a+b-3a+3b)[a^2+2ab+b^2+3a^2-3b^2+9(a^2-2ab+b^2)]
=(4b-2a)(13a^2-16ab+7b^2)
=2(2b-a)(13a^2-16ab+7b^2)
收錄日期: 2021-04-25 17:17:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080103000051KK03840