Eg.11 a) Expand in ascending powers of y as far as y^3
i) (1-y+3y^2)^3
ii)(1-y+3y^2-12y^3)^3
b) Using the results in a), show that if powers of x above the third are
neglected, an approximate solution of the cubic equation
yx^3+x-1=0 is x=1-x+3x^2-12x^3
c) Hence find an approximate solution of the cubic equation x^3+10x-10=0.
Binomial expression
2008-07-03 11:17 pm
回答 (2)
2008-07-04 12:06 am
✔ 最佳答案
(1-y+3y^2)^3=[1-y(1+3y)]^3
=(1)(1)-3(y)(1+3y)+3(y^2)(1+3y)^2-(y^3)(1+3y)^3
=1-3y+12y^2-19y^3+....//
2008-07-03 16:07:08 補充：
(1-y+3y^2-12y^3)^3
=1-3y+12y^2-55y^3+....//
2008-07-07 19:08:26 補充：
(1-y+3y^2-12y^3)^3
=-1728y^9+1296y^8-756y^7+675y^6-279y^5+108y^4-55y^3+12y^2-3y+1
2008-07-04 4:19 am
GOOD
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https://hk.answers.yahoo.com/question/index?qid=20080703000051KK01352