Question

Find the squares of $99$ using the identity
$(a-b{)}^2=a^2-2ab+b^2$

Answer 1

${99}^2={(100-1)}^2$
It is the form of ${(a-b)}^2$, where $a=100,b=1$
Applying the formula ${(a-b)}^2=a^2+b^2-2ab$
${99}^2={(100-1)}^2={100}^2+1^2-2\times 100\times 1=10000+1-200=9801$