Variables, Integers

C. 2

Since $a, b, c$ are consecutive integers
$$
\Rightarrow a=b-1 \text { and } c=b+1
$$
Expression: $\frac{a^3+b^3+c^3+3 a b c}{(a+b+c)^2}$
$$
\begin{aligned}
& =\frac{(b-1)^3+b^3+(b+1)^3+3(b-1) b(b+1)}{(b-1+b+b+1)^2} \\
& =\frac{b^3+3 b+b^3+b^3+3 b+3 b^3-3 b}{9 b^2} \\
& =\frac{6 b^3+3 b}{9 b^2}=\frac{2 b^2+1}{3 b}
\end{aligned}
$$
Putting different values of $b$ from – 10 to 10 , we can verify that only – 1 and 1 satisfies to get integer values for the expression.