[(a+b)(a-b)]^3
=
[(a+b)(a-b)]^3=
2006-11-12 5:23 am
回答 (7)
2006-11-14 12:17 am
[(a+b)(a-b)]^3
=(a^2-b^2)^3
=a^6-b^6
=(a-b)(a+b)(a-b)(a+b)(a-b)(a+b)
=(a+b)^3(a-b)^3
=(a^3+3ab^2+3ba^2+b^3)
=a^3+3ab^2-3a^2b-b^3
=(a^2-b^2)^3
=a^6-b^6
=(a-b)(a+b)(a-b)(a+b)(a-b)(a+b)
=(a+b)^3(a-b)^3
=(a^3+3ab^2+3ba^2+b^3)
=a^3+3ab^2-3a^2b-b^3
參考: me
2006-11-12 6:04 am
consider the polynomial,
(A+B)^3 = A^3 + 3A^2 * B + 3AB^2 + B^3
[(a+b)(a-b)]^3
= (a^2 - b^2)^3
PUTTING A=a^2 and B= -b^2 into the polynomial, then
(a^2 - b^2)^3
= a^2^3 + 3a^2^2 * (-b^2) + 3a^2 * (-b^2)^2 + (-b^2)^3
= a^6 - 3a^4 * b^2 + 3a^2 * b^4 - b^6
(A+B)^3 = A^3 + 3A^2 * B + 3AB^2 + B^3
[(a+b)(a-b)]^3
= (a^2 - b^2)^3
PUTTING A=a^2 and B= -b^2 into the polynomial, then
(a^2 - b^2)^3
= a^2^3 + 3a^2^2 * (-b^2) + 3a^2 * (-b^2)^2 + (-b^2)^3
= a^6 - 3a^4 * b^2 + 3a^2 * b^4 - b^6
2006-11-12 5:46 am
[(a+b)(a-b)]^3
=(a^2-b^2)^3
By binomial theorem
=(a^2)^3 + [3(a^2)^2](- b^2) + 3(a^2)[(- b^2)^2] + (- b^2)^3
=a^6 - 3(a^4)(b^2) + 3(a^2)(b^4) - b^6
=(a^2-b^2)^3
By binomial theorem
=(a^2)^3 + [3(a^2)^2](- b^2) + 3(a^2)[(- b^2)^2] + (- b^2)^3
=a^6 - 3(a^4)(b^2) + 3(a^2)(b^4) - b^6
2006-11-12 5:29 am
use binomial formula
[(a+b)(a-b)]^3
= (a^2-b^2)^3
= (a^2)^3+3(a^2)^2(-b^2)+3(a^2)(-b^2)^2+(-b^2)^3
=a^6-3a^4b^2+3a^2b^4-b^6
[(a+b)(a-b)]^3
= (a^2-b^2)^3
= (a^2)^3+3(a^2)^2(-b^2)+3(a^2)(-b^2)^2+(-b^2)^3
=a^6-3a^4b^2+3a^2b^4-b^6
2006-11-12 5:26 am
=[a^2-b^2]^3
=a^6-3a^4b^2+3a^2b^4-b^6
=a^6-3a^4b^2+3a^2b^4-b^6
2006-11-12 5:25 am
[(a+b)(a-b)]^3
= -9
= -9
收錄日期: 2021-04-12 22:56:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061111000051KK05246