We rewrite the expression as $\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}$. If $a=b$ then the top is $a^3(a-c)-a^3(a-c)-c^3(a-a)=0$, and so $a-b$ is a factor of the top. Similarly, $a-c$ and $b-c$ are too. Thus the expression is $\frac{k(a+b+c)(a-b)(b-c)(a-c)}{(a-b)(b-c)(a-c)}=k(a+b+c)$, because the original expression is symmetric in $a, b, c$ and $k(a+b+c)$ is the only degree-1 symmetric expression. Then let $a=1, b=2, c=0$ to get $\frac{1}{-1}+\frac{8}{2}+0=3=a+b+c$, so $k=1$. Thus the expression is $a+b+c$, and so clearly $a+b+c$ is an integer iff $\{a\}+\{b\}+\{c\}$ is.