Question

One of the factors of $(a^2-b^2)(c^2-d^2)-4abcd$ is ______.

• A. $(ac-bd+bc+ad)$
• B. $(ac-bd+bc-ad)$
• C. $(ac+bd+bc-ad)$
• D. $(ac+bd+bc+ad)$

Answer 1

The correct option is A $(ac-bd+bc+ad)$

$(a^2-b^2)(c^2-d^2)-4abcd=a^2{c}^2-a^2{d}^2-b^2{c}^2+b^2{d}^2-4abcd$
$=a^2{c}^2-2abcd+b^2{d}^2-a^2{d}^2-2abcd-b^2{c}^2$
$=(ac{)}^2-2(ac\left)\right(bd)+(bd{)}^2-\left[\right(ad{)}^2+2\left(ad\right)\left(bc\right)+\left(bc{)}^2\right]$

Using identity:
$(a+b{)}^2=a^2+2ab+b^2\text{and}(a-b{)}^2=a^2-2ab+b^2$
we get,
$=(ac-bd{)}^2-(ad+bc{)}^2$
we know that, $a^2-b^2=(a+b)(a-b)$
$\Rightarrow (ac-bd{)}^2-(ad+bc{)}^2=(ac-bd+bc+ad)(ac-bd-bc-ad)$

Therefore, $(ac-bd+ad+bc)and(ac-bd-ad-bc)$
are the factors of $(a^2-b^2)(c^2-d^2)-4abcd.$