Question
Simplify the following using the identities: $\frac{198\times 198−102\times 102}{96}$
✓
Answer
Let us consider the following expression:
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$ Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$
$=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{300\times 96}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=300$
Thus, the answer is $300.$
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$ Using the identity $(a+b)(a-b)=a^2-b^2$
we get:
$\frac{198\times 198−102\times 102}{96}=\frac{198^2-102^2}{96}$
$=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{(198+102)(198−102)}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=\frac{300\times 96}{96}$
$\Rightarrow\frac{198\times 198−102\times 102}{96}=300$
Thus, the answer is $300.$
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