Question

(a square minus b) whole
square into (c square minus d)whole square minus 4 ABCD

Answer 1

(ac-bd+ad+bc)(ac-bd-ad-bc)

Answer 2

$(a^{2}-b^{2})(c^{2}-d^{2}) - 4abcd\\= a^{2}(c^{2}-d^{2}) - b^{2}(c^{2}-d^{2}) - 4abcd \\= a^{2}c^{2} - a^{2}d^{2} - b^{2}c^{2} + b^{2}d^{2} - 4abcd$

/* Rearranging the terms, we get */

$= a^{2}c^{2} + b^{2}d^{2} - 2abcd - a^{2}d^{2} - b^{2}c^{2} - 2abcd$

$=[ (ac)^{2} + (bd)^{2} - 2\times ac\times bd] -[ (ad)^{2} + (bc)^{2} - 2\times ad \times bc]$

$= (ac-bd)^{2} - (ad-bc)^{2} \\= [(ac-bd)+(ad-bc)][(ac-bd)-(ad-bc)]$

$\blue {(By \: Algebraic\: Identity )}$

$\boxed {\pink {x^{2} - y^{2} = ( x + y )(x - y ) }}$

$= (ac-bd+ad-bc)(ac-bd-ad+bc)$

Therefore.,

$\red{(a^{2}-b^{2})(c^{2}-d^{2}) - 4abcd}$

$\green {= (ac-bd+ad-bc)(ac-bd-ad+bc)}$

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