Algebraic Expression Integer Values

Let $a = b - 1$, $b = b$, $c = b + 1$. Sum $a + b + c = 3b$. $a^3 + b^3 + c^3 + 3abc = (b-1)^3 + b^3 + (b+1)^3 + 3(b-1)b(b+1) = 6b^3 + 3b$. Denominator = $(3b)^2 = 9b^2$. Expression = $\frac{6b^3 + 3b}{9b^2} = \frac{2b^2 + 1}{3b}$. For this to be an integer, $3b$ must divide $2b^2 + 1$. Testing non-zero $b \in [-9, 9]$: If $b = 1 \implies \frac{3}{3} = 1$. If $b = -1 \implies \frac{3}{-3} = -1$. No other integers $b$ make $\frac{2b^2 + 1}{3b}$ an integer. Thus, exactly 2 integer values are possible.