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
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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,

\[ 104\times96=9984 \]

 

 

 

 

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