What is the value of $\displaystyle\frac{109^2-100^2}{9}$?
I notice that the numerator of the fraction is the difference of two perfect squares, so I can use the identity $(a^2-b^2)=(a+b)(a-b)$ to factor it. In this case, $a=109$ and $b=100$, so I get $\displaystyle\frac{109^2-100^2}{9}=\frac{(109+100)(109-100)}{9}$. This simplifies to $\displaystyle\frac{209\cdot 9}{9}$, which is just $209$. 