Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9$ as we only care $\pmod 9.$
So $(x+a)(x+b)(x+c)\equiv 1,2,0\mod 9.$
Note that $$\frac{\prod_{x=0}^8 (x+a)}{(a+2)(a+5)(a+8)}\cdot  \frac{\prod_{x=0}^8(x+b)}{(b+2)(b+5)(b+8)}\cdot \frac{\prod_{x=0}^8(x+c)}{(c+2)(c+5)(c+8)}= [1\cdot 2\cdot 4\cdot 5\cdot 7\cdot 8]^3\equiv -1^3\equiv -1 $$
I am stuck. Solutions are appreciated.

I'd start by proving some algebraic relations between $a,b,c$ in $\mathbb{Z}/9\mathbb{Z}.$ For instance, by plugging in $x=0$ we have
$$abc\equiv 1 \pmod 9$$
so each of $a,b,c$ is a unit: congruent to either $1,2,4,5,7$ or $8$. It follows that the square of each of $a,b,c$ is either $1,4$ or $7$ (mod $9$).
Expanding $A$ and using the above relation we have
$$x^2(a+b+c) + x(ab+ac+bc) \equiv 0 \pmod 9$$
and plugging in $x=1$ and $x=-1$ and summing we get
$$a+b+c\equiv 0 \pmod 9$$
so that
\begin{align*}
x(ab + ac + bc) &\equiv 0 \pmod 9\\
x ((a+b+c)^2 - a^2-b^2-c^2) &\equiv 0 \pmod 9\\
x(a^2+b^2+c^2) &\equiv 0 \pmod 9.
\end{align*}
Plug in $x=1$ to get $a^2+b^2+c^2\equiv 0 \pmod 9$.
Well, the only combination of $\{1,4,7\}$ that sums to zero mod 9 is (WLOG) $a^2\equiv 7, b^2\equiv 1,c^2\equiv 1.$ But then $a\equiv \pm 4$ and $b,c\equiv \pm 1$ and no combination of signs will give you $abc\equiv 1.$
