(a) x to the power 9 - 1
(b) x to the power 6 - 64
請問點樣可以清楚例出步驟計算?
sum and difference of two cubes
2007-03-04 7:43 pm
回答 (2)
2007-03-04 8:21 pm
✔ 最佳答案
利用立方差的公式 x3 – y3 = (x – y)(x2 + xy + y2)
(a) x9 - 1
= (x3)3 - 1
= (x3 – 1)[(x3)2 + x3 + 1]
= (x – 1)(x2 + x + 1)(x6 + x3 + 1)
(b) x6 – 64
= (x2)3 – 43
= (x2 – 4)[(x2)2 + 4x2 + 42]
= (x – 2)(x + 2)(x4 + 4x2 + 16)
2007-03-04 8:00 pm
x³士y³=(x士y)(x²干xy+y²)
proof:
R.H.S.
=(x士y)(x²干xy+y²)
=x³干x²y+xy²士x²y-xy²士y³
=x³干x²y士x²y+xy²-xy²士y³
=x³士y³
The proof is completed.
(a)
x9-1
=(x3)3-(1)3
=(x3-1)(x6-x3+1)
(b)
x6-64
=(x2)3-43
=(x2-3)(x4-4x2+16)
2007-03-04 17:54:15 補充：
sorry,x^6-64=(x^2)^3-4^3=(x^2-4)(x^4-4x^2十16)=(x十2)(x-2)(x^4-4x^2十16)
proof:
R.H.S.
=(x士y)(x²干xy+y²)
=x³干x²y+xy²士x²y-xy²士y³
=x³干x²y士x²y+xy²-xy²士y³
=x³士y³
The proof is completed.
(a)
x9-1
=(x3)3-(1)3
=(x3-1)(x6-x3+1)
(b)
x6-64
=(x2)3-43
=(x2-3)(x4-4x2+16)
2007-03-04 17:54:15 補充：
sorry,x^6-64=(x^2)^3-4^3=(x^2-4)(x^4-4x^2十16)=(x十2)(x-2)(x^4-4x^2十16)
收錄日期: 2021-04-23 00:10:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070304000051KK01423