(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)