Question

请问（a+b+c）³-（a³+b³+c³）分解因式怎么分？

Answer 1

解（a+b+c）³-（a³+b³+c³）
=（a+b+c）³-a³-（b³+c³）
=[（a+b+c）-a][（a+b+c）^2+（a+b+c）a+a^2]-(b+c)[b^2-bc+c^2]
=(b+c)[a^2+b^2+c^2+2ab+2bc+2ca+a^2+ab+ac+a^2]-(b+c)(b^2-bc+c^2)
=(b+c)[a^2+b^2+c^2+2ab+2bc+2ca+a^2+ab+ac+a^2-(b^2-bc+c^2)]
=(b+c)[3a^2+3ab+3ac+3bc]
=(b+c)[3a(a+b)+3c(a+b)]
=3(b+c)(a+b)(a+c)

Answer 2

（a+b+c）³-（a³+b³+c³）=a³+b³+c³+3a²b+3a²c+6abc+3ab²+3ac²+3b²c+3bc²-（a³+b³+c³）
=3a²b+3a²c+6abc+3ab²+3ac²+3b²c+3bc²
=3a²(b+c)+6abc+3a(b²+c²)+3bc(b+c)
=(b+c)(3a²+3bc)+3a(b²+c²+2bc)
=3(b+c)(a²+bc+ab+ac)
=3(b+c)[a(a+b)+c(a+b)]
=3(b+c)(a+b)(a+c)

Answer 3

(a+b+c)^3-(a^3+b^3+c^3)
=(a+b+c)^3-[a^3+b^3+c^3-3abc]-3abc
=(a+b+c)^3-[(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)]-3abc
=(a+b+c)^3-{[(a+b)^3+c^3]-3ab(a+b+c)}-3abc
=(a+b+c)^3-[(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)]-3abc
=(a+b+c)^3-(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)-3abc
=(a+b+c)^3-(a+b+c)(a^2+b^2+c^2-ab-bc-ac)-3abc
=(a+b+c)[(a+b+c)^2-(a^2+b^2+c^2-ab-bc-ac)]-3abc
=(a+b+c)[a^2+b^2+c^2+2ab+2ac+2bc-a^2-b^2-c^2+ab+bc+ac)]-3abc
=(a+b+c)(3ab+3ac+3bc)-3abc
=3[(a+b+c)(ab+ac+bc)-abc]