Step-by-step explanation:
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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Verified answer
(a² - b²)( c² - d²) - 4abcd= {a²c² -a²d² -b²c² + b²d² } -4abcd
= { a²c² + b²d² - 2abcd} -{ a²d² + b²c² +2abcd}
= { (ac)² + (bd)² -2(ac)(bd)} - {(ad)² + (bc)² +2( ad)( bc) }
= ( ac - bd)² - (ad +bc)²
= ( ac - bd + ad + bc)( ac -bd -ad - bc)
Step-by-step explanation:
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)]