Problem (Algebra):
Let $a=19, b=20$, and $c=21$. Compute

$$
\frac{a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a}{a+b+c}
$$

Solution:
Note that the numerator of the given expression factors as $(a+b+c)^{2}$, hence the expression to be computed equals $a+b+c=19+20+21=\mathbf{6 0}$.

Final answer: 60