Question

分解因式(a+b+c+d)(b+c-a-d)(c+a-b-d)(a+b-c-d)+16abcd

Answer 1

(注：^代表平方，打不出来啊) 解:原式=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd =[(a+b)^-(c+d)^][(c-d)^-(b-a)^]+16abcd =(a^+b^-c^-d^+2ab-2cd)(c^+d^-a^-b^-2cd+2ab)+16abcd =(2ab-2cd)^-(a^+b^-c^-d^)^+16abcd =4a^b^+4c^d^-8abcd+16abcd-(a^+b^-c^-d^) =4a^b^+4c^d^+8abcd-(a^+b^-c^-d^) =(2ab+2cd)^-(a^+b^-c^-d^)^ =(2ab+2cd+a^+b^-c^-d^)(2ab+2cd-a^-b^+c^+d^) =...... =(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) (思路：前面4个等号都是用平方差公式，第5等号解决掉+16abcd，之后基本是照原路返回)^_^