Question

分解因式(a-b)³+(b-c)³-(a-c)³

Answer 1

首先推导公式（x-y)^3=x^3-3(X^2)y+3x(Y^2)-y^3

(a-b)^3=a^3-3a^2b+3cb^2-b^3   ……1

(b-c)^3=b^3-3b^2c+3bc^2-c^3…………2

(a-c)^3=a^3-3a^2c+3ac^2-c^3…………3

因此=(a-b)³+(b-c)³-(a-c)³=

Answer 2

原式=[(a-b)3+(b-c)3]-(a-c)3
=[(a-b)+(b-c)][(a-b)2-(a-b)(b-c)+(b-c)2]-(a-c)3
=(a-c)[(a-b)2-(a-b)(b-c)+(b-c)2]-(a-c)3
=(a-c)[(a-b)2-(a-b)(b-c)+(b-c)2-(a-c)2]
=(a-c){(a-b)2-(a-b)(b-c)+[(b-c)+(a-c)][(b-c)-(a-c)]
=(a-c){(a-b)2-(a-b)(b-c)+(b+a-2c)(b-a)}
=(a-c){(a-b)(a-2b+c)-(a-b)(b+a-2c)}
=(a-c){(a-b)(a-2b+c-b-a+2c)
=3(a-c)(a-b)(c-b)

括号后面的数字均为次方，括号前面的是普通数字。应该能看懂吧，页面没法弄出来次方的格式

Answer 3

(a-b)³+(b-c)³-(a-c)³
=--3 (a - b) (a - c) (b - c)

Answer 4

式(a-b)³+(b-c)&