原式=(a+b)^3-3(a+b)^2*c+3(a+b)*c^2-c^3-(a^3+b^3+c^3)
=(a^3+b^3+3a^2b+3ab^2)-3(a^2+2ab+b^2)*c+3ac^2+3bc^2-c^3-(a^3+b^3+c^3)
=a^3+3a^2b+3ab^2+b^3-3a^2c-6abc-3b^2c+3ac^2+3bc^2-a^3-b^3-2c^3
=3a^2b+3ab^2-3a^2c-6abc-3b^2c+3ac^2+3bc^2-2c^3
=3a^2(b-c)+3a(b^2-2bc+c^2)-3b^2c+3bc^2-2c^3
=3a^2(b-c)+3a(b-c)^2-3bc(b-c)-2c^3
=3(b-c)(a^2+ab-ac-bc)-2c^3
后面不好變換了
如果前面的括號里的c的系數為正,本題結果為
3(a+b)(a+c)(b+c)
題目可能有點問題.