Question

Factorise:
(ac-bd)²-(ad-bc)²​

Answer 1

Answer:

given : (a²−b²)(c²−d²)−4abcd

=a²c²−a²d²−b²c²+b²d²−2abcd−2abcd

=a²c²+b²d²−2abcd−a²d²+b²c²−2abcd

=(ac−bd)²−(ad−bc)²

=(ac−bd−ad+bc)(ac−bd+ad−bc) ....since [p²−q²=(p−q)(p+q)]

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