點因式分解
a^3+b^3+3ab(a+b)
要有過程
因式分解問題
2008-01-10 4:26 am
回答 (2)
2008-01-10 4:30 am
✔ 最佳答案
a^3+b^3+3ab(a+b)=(a+b)(a^2-ab+b^2)+3ab(a+b)
=(a+b)(a^2-ab+3ab+b^2)
=(a+b)(a^2+2ab+b^2)
=(a+b)(a+b)^2
=(a+b)^3
2008-01-10 5:02 am
a^3+b^3+3ab(a+b)
=(a^3+b^3)+3ab(a+b)
=(a+b)(a^2-ab+b^2)+3ab(a+b)
=(a+b)(a^2-ab+b^2+3ab)
=(a+b)(a^2+2ab+b^2)
=(a^3+b^3)+3ab(a+b)
=(a+b)(a^2-ab+b^2)+3ab(a+b)
=(a+b)(a^2-ab+b^2+3ab)
=(a+b)(a^2+2ab+b^2)
收錄日期: 2021-04-13 14:54:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080109000051KK03600