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Question

Factorise :
$$(a^2-1)(b^2-1)+4ab$$


Solution

$$(a^2-1)(b^2-1)+4ab=a^2b^2-a^2-b^2+1+4ab$$
$$(a^2-1)(b^2-1)+4ab=a^2b^2+1+2ab-a^2-b^2+2ab$$
$$(a^2-1)(b^2-1)+4ab=(a^2b^2+1+2ab)-(a^2+b^2-2ab)$$
$$(a^2-1)(b^2-1)+4ab=(ab+1)^2-(a-b)^2$$  $$[\therefore a^2-b^2=(a+b)(a-b)]$$
$$(a^2-1)(b^2-1)+4ab=(ab+1-a+b)(ab+1+a-b)$$

Mathematics

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