Question

( a²-b²)(c²-d²)-4abcd factories it

Answer 1

Answer:

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

Answer 2

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Factorise

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

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We have,

$\color{brown} \small{ \bf({a}^{2} - {b}^{2} )( {c}^{2} - {d}^{2} ) - 4abcd} \\ \\ \color{brown}\small{ \bf {a}^{2} {c}^{2} - {a}^{2} {d}^{2} - {b}^{2} {c}^{2} + {b}^{2} {d}^{2} - 4abcd} \\ \\ \color{brown}\small{ \bf {(ac)}^{2} - {(ad)}^{2} - {(bc)}^{2} + {(bd)}^{2} - 4abcd }$

Rearranging the terms

$\color{brown}\small{ \bf {(ac)}^{2} + {(bd)}^{2} - {(ad)}^{2} - {(bc)}^{2} - 4abcd }$

Taking

- 4abcd = - 2 abcd + ( - 2abcd)

$\color{brown}\small \bf {(ac)}^{2} + {(bd)}^{2} - 2abcd - {(ad)}^{2} - {(bc)}^{2} - 2abcd \\ \\ \color{brown} \small \bf {(ac)}^{2} + {(bd)}^{2} - 2(ac)(bd) - ( {(ad)}^{2} + {(bc)}^{2} + 2(ad)(bc))$

Using identity

$\tiny \tt {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2}$

and

$\tiny \tt {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2}$

We will get

$\color{brown} \small \bf {(ac - bd)}^{2} - ( {(ad + bc)}^{2} )$

Using identity

$\tiny \tt {x}^{2} - {y}^{2} = (x + y)(x - y)$

We will get

$\boxed{\small\bf \color{brown}(ac - bd + ad + bc) \: (ac - bd - ad - bc)}$

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