Question

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

• A. $(1+ab+a-b)(1+ab+a+b)$
• B. $(1+ab+a-b)(1+ab-a^2+b)$
• C. $(1+a{b}^2+a-b)(1+ab-a+b)$
• D. $(1+ab+a-b)(1+ab-a+b)$

Answer 1

The correct option is D $(1+ab+a-b)(1+ab-a+b)$

$(1-a^2)(1-b^2)+4ab$
$=1-a^2-b^2+a^2{b}^2+2ab+2ab$
$=1+2ab+a^2{b}^2-a^2+2ab-b^2$
$=[1+2ab+(ab{)}^2]-[a^2-2ab+b^2]$
We know that,
{ $a^2-b^2=(a+b)(a-b)$}
$=(1+ab{)}^2-(a-b{)}^2$
$=\left[\right(1+ab)+(a-b\left)\right]\left[\right(1+ab)-(a-b\left)\right]$
$=(1+ab+a-b)(1+ab-a+b)$