Question

(ab+cd)^2-(ac+bd)^2 因式分解

Answer 1

=(ab+cd+ac+bd)(ab+cd-ac-bd)

=[(ab+ac)+(cd+bd)][(ab-ac)+(cd-bd)]

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

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