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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