how to expand (2a-3b)^3 using the binomial thereom?
2010-01-11 11:40 am
can someone please do this queston for me and show me all the steps please! thanks :)
回答 (3)
2010-01-11 12:20 pm
✔ 最佳答案
(2a-3b)^3recall the pascals triangle in th link:
http://www.flickr.com/photos/anoved/2290765890/
we use raw 3 whose coefficient is 1 3 3 1
or (x+y)^3=x^3+3x^2y+3xy^2+y^3
(2a-3b)^3
=(2a)^3+3(2a)^2(-3b)+3(2a)(-3b)^2+(-3b)^3
=8a^3-36a^2b+54ab^2-27b^3 answer//
2010-01-11 7:59 pm
(x-y)^3 = x^3 - 3*(x^2)*y + 3*x*(y^2) - y^3
now all you have to do is replace x and y by 2a and 3b :
(2a - 3b)^3 = (2a)^3 - 3*(2a)^2*(3b) + 3*(2a)*(3b)^2 - (3b)^3 = 8a^3 - 36(a^2)*b + 54a*(b^2) - 27b^3
now all you have to do is replace x and y by 2a and 3b :
(2a - 3b)^3 = (2a)^3 - 3*(2a)^2*(3b) + 3*(2a)*(3b)^2 - (3b)^3 = 8a^3 - 36(a^2)*b + 54a*(b^2) - 27b^3
2010-01-11 7:51 pm
(2a - 3b)^3
= C(3, 0)(2a)^3(-3b)^0 + C(3, 1)(2a)^2(-3b)^1 + C(3, 2)(2a)^1(-3b)^2 + C(3, 3)(2a)^0(-3b)^3
= 3!/(0!3!)(8a^3) + 3!/(1!2!)(4a^2)(-3b) + 3!/(2!1!)(2a)(9b^2) + 3!/(3!0!)(-27b^3)
= 8a^3 - 36a^2b + 54ab^2 - 27b^3
NB:
x! = x(x - 1)(x - 2)(x - 3) ... (x - x + 1)
C(x, y) = x!/[y!(x - y)!]
= C(3, 0)(2a)^3(-3b)^0 + C(3, 1)(2a)^2(-3b)^1 + C(3, 2)(2a)^1(-3b)^2 + C(3, 3)(2a)^0(-3b)^3
= 3!/(0!3!)(8a^3) + 3!/(1!2!)(4a^2)(-3b) + 3!/(2!1!)(2a)(9b^2) + 3!/(3!0!)(-27b^3)
= 8a^3 - 36a^2b + 54ab^2 - 27b^3
NB:
x! = x(x - 1)(x - 2)(x - 3) ... (x - x + 1)
C(x, y) = x!/[y!(x - y)!]
收錄日期: 2021-05-01 12:56:37
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