Question

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

Answer 1

原式=[(a+b)+(c+d)][(c-d)+(b-a)][(c-d)-(b-a)][(a+b)-(c+d)]+16abcd

=[(a+b)^2-(c+d)^2][(c-d)^2-(b-a)^2]+16abcd

=(a^2+b^2-c^2-d^2+2ab-2cd)(c^2+d^2-a^2-b^2-2cd+2ab)+16abcd

=(2ab-2cd)^2-(a^2+b^2-c^2-d^2)^2+16abcd

=4a^2b^2+4c^2d^2-8abcd+16abcd-(a^2+b^2-c^2-d^2)

=4a^2b^2+4c^2d^2+8abcd-(a^2+b^2-c^2-d^2)

=(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2

=(2ab+2cd+a^2+b^2-c^2-d^2)(2ab+2cd-a^2-b^2+c^2+d^2)

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

=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)