Question

(a²-b²)(c²-d²)-4abcd
Factorise it.

Chance to get 100 points.
Any fake answers will be reported.

Answer 1

: (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)]

Answer 2

: (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)]

Answer 3

: (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)]

Answer 4

: (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)]

Answer 5

: (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)]