Question

(a+b+c)³-(a³+b³+c³） 怎么因式分解

Answer 1

(a+b+c)³-(a³+b³+c³）

=(a+b+c)³-a³-(b³+c³)

=(a+b+c-a)(a+b+c)^2+a(a+b+c)+a^2-(b+c)(b^2-bc+c^2)

=(b+c)[(a+b+c)^2+a(a+b+c)+a^2-(b^2-bc+c^2)]

=(b+c)[3a^2+3ab+3bc+3ac]

=3(b+c)[a(a+b)+c(a+b)]

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