Question

factorisation = (a²-b²-c²+d²)-4(ad-bc) ²​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

$( {a}^{2} - {b}^{2} - {c}^{2} + {d}^{2} ) - 4 {(ad - bc)}^{2}$

Identity used here :-

$\bold{\boxed{ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}$

$= > ( {a}^{2} - {b}^{2} - {c}^{2} + {d}^{2} ) - 4( {a}^{2} {d}^{2} + {b}^{2} {c}^{2} - 2adbc)$

$= > {a}^{2} - {b}^{2} - {c}^{2} + {d}^{2} - 4 {a}^{2} {d}^{2} - 4 {b}^{2} {c}^{2} + 8adbc$

$= > {a}^{2} - 4 {a}^{2} {d}^{2} - {b}^{2} - 4 {b}^{2} {c}^{2} - {c}^{2} + {d}^{2} + 8adbc$

$= > ( {a}^{2} - 4 {a}^{2} ) {d}^{2} - ( {b}^{2} + 4 {b}^{2} ) {c}^{2} - {c}^{2} + {d}^{2} + 8adbc$

$= > ( - 3 {a}^{2} ) {d}^{2} - (5 {b}^{2} ) {c}^{2} - ( - c - d + 8ab)cd$

$\bold{= > ( - 3 {a}^{2} ) {d}^{2} - ( 5{b}^{2} ) {c}^{2} - (8ab - c - d)cd}$

$\bold\fcolorbox{aqua}{orange}{Answer}$

$\bold{( - 3 {a}^{2} ) {d}^{2} - ( 5{b}^{2} ) {c}^{2} - (8ab - c - d)cd}$