factorise (2x+1)^3 + (2x-1)^3?

2009-02-14 11:14 am

更新1:

okay i agree but a³ + b³ = (a+b)(a²-ab+b²) (2x+1)³+(2x-1)³ SHOULDNT THE BELOW STEP BE ((2x+1) + (2x-1))((2x+1)² - (2x+1)(2x-1) + (2x-1)²) therefore 2x +1 +2x +1?? (and the rest....) = (2x+1 + 2x-1)((2x+1)² - (2x+1)(2x-1) + (2x-1)²) = 4x(4x²+4x+1-4x²+1+4x²-4x+1) = 4x(4x²+3)

更新2:

回答 (4)

[匿名]

2009-02-14 11:56 am

✔ 最佳答案

(2x+1)^3
=8x^3+12x^2+6x+1......(1)
(2x-1)^3
=8x^3-12x^2+6x-1........(2)

add (1) and (2)

(2x+1)^3+(2x-1)^3
=16x^3+12x
=4x(4x^2+3)

👍 0 👎 0

gudspeling

2009-02-14 7:22 pm

aÂ³ + bÂ³ = (a+b)(aÂ²-ab+bÂ²)

(2x+1)Â³+(2x-1)Â³
= (2x+1 + 2x-1)((2x+1)Â² - (2x+1)(2x-1) + (2x-1)Â²)
= 4x(4xÂ²+4x+1-4xÂ²+1+4xÂ²-4x+1)
= 4x(4xÂ²+3)

👍 2 👎 0

An ESL Learner

2009-02-14 8:26 pm

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

(2x + 1)^3 + (2x - 1)^3
= [(2x + 1) + (2x - 1)][(2x + 1)^2 - (2x + 1)(2x - 1) + (2x - 1)^2]
= [2x + 1 + 2x - 1][(2x + 1)(2x + 1) - (4x^2 + 2x - 2x - 1) + (2x - 1)(2x - 1)]
= [2x + 2x + 1 - 1][4x^2 + 2x + 2x + 1 - 4x^2 + 1 + 4x^2 - 2x - 2x + 1]
= 4x[4x^2 + 4x + 1 - 4x^2 + 1 + 4x^2 - 4x + 1]
= 4x[4x^2 - 4x^2 + 4x^2 + 4x - 4x + 1 + 1 + 1]
= 4x[4x^2 + 3]

👍 0 👎 0

[匿名]

2009-02-14 7:23 pm

(2x+1)^3 + (2x-1)^3
= [(2x+1) + (2x-1)]*[(2x+1)^2- (2x+1)(2x-1)+(2x-1)^2]

👍 0 👎 0

收錄日期: 2021-05-01 12:06:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090214031431AAw6YzR

檢視 Wayback Machine 備份