d Problem: If a^2-b^2=(a+b)(a-b), then find the value of 104 x 96. | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Problem: If a^2-b^2=(a+b)(a-b), then find the value of 104 x 96.

Problem: If ​$$a^2-b^2=(a+b)(a-b)$$​ then find the value of 104 x 96.

Solution:

We are given that;

$a^2-b^2=(a+b)(a-b)$

or

$(a+b)(a-b)=a^2-b^2$

So, to find the value of ​$$104\times96$$​ with reference to the given formula we first arrange it in the form of formula as below;

$104\times96=(100+4)(100-4)$

Here LHS i-e ​$$(100+4)(100-4)$$​is in the form of ​$$(a+b)(a-b)$$​, which is equal to ​$$a^2-b^2$$

So,

$104\times96=(100+4)(100-4)=100^2-4^2=1000-16=9984$

Hence we can say,

$104\times96=9984$