One hard question:
Please answer the following question with CLEAR STEPS.
(3a-4b)[[^5]] - 72a[[^2]]b[[^3]] + 192ab[[^4]] - 128b[[^5]]
PS: [[^5]]= to the power of 5, [[^2]]= square, [[^3]]= cube, [[^4]]= to the power of 4, 只是想清楚一點！
Factorization using other identities
2008-12-10 4:06 am
回答 (2)
2008-12-10 4:44 am
✔ 最佳答案
(3a-4b)^5 - 72 a^2 b^3 + 192 a b^4 - 128 b^5=(3a-4b)^5 - 8 b^3 [9a^2-24ab+16b^2]
=(3a-4b)^5 -(2b)^3 (3a-4b)^2
=(3a-4b)^2 [(3a-4b)^3 - (2b)^3]
=(3a-4b)^2 [(3a-4b) - (2b)][(3a-4b)^2 + (2b)(3a-4b) + (2b)^2]
=(3a-4b)^2 [3a-6b][9a^2 - 18ab +12b^2]
=(3a-4b)^2*3(a-2b)*3(3a^2 - 6ab + 4b^2)
=9 (3a-4b)^2 (a-2b) (3a^2 - 6ab + 4b^2)
2008-12-09 20:52:36 補充：
http://lrg103.zorpia.com/0/4927/31536658.1534f4.jpg
參考: ME
2008-12-10 4:42 am
(3a - 4b)5 - 72a2b3 + 192ab4 - 128b5
= (3a - 4b)5 - 8b3(9a2 - 24ab - 16b2)
= (3a - 4b)5 - 8b3[(3a)2 - 2(3a)(4b) + (4b)2]
= (3a - 4b)5 - 8b3(3a - 4b)2
= (3a - 4b)2[(3a - 4b)3 - 8b3]
= (3a - 4b)2[(3a - 4b)3 - (2b)3]
= (3a - 4b)2[(3a - 4b) - 2b][(3a - 4b)2 + (3a - 4b)(2b) + (2b)2]
= (3a - 4b)2(3a - 4b - 2b)(9a2 - 24ab + 16b2 + 3ab - 8b2 + 4b2)
= (3a - 4b)2(3a - 6b)(9a2 - 18ab + 12b2)
= 3(3a - 4b)2(3a - 6b)(3a2 - 6ab + 4b2)
=
收錄日期: 2021-04-25 13:39:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081209000051KK01373