factorise x^6-1
請問應該點計?.?
列埋d step比我哇
吾該哂!!!!!!!
f.2 maths一問
2008-06-13 7:21 pm
回答 (2)
2008-06-13 8:04 pm
x^6-1
=(x^3+1)(x^3-1)
=(x+1)(x^2-x+1)(x-1)(x^2+x+1)
=(x+1)(x-1)(x^2-x+1)(x^2+x+1)
=(x^3+1)(x^3-1)
=(x+1)(x^2-x+1)(x-1)(x^2+x+1)
=(x+1)(x-1)(x^2-x+1)(x^2+x+1)
2008-06-13 7:50 pm
x^6-1
= (x^3-1)(x^3+1)
= (x-1)(x^2+x+1)(x^3+1)
= (x^3-1)(x^3+1)
= (x-1)(x^2+x+1)(x^3+1)
收錄日期: 2021-04-20 22:56:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080613000051KK00589