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.

Perhaps a slight hint should do good, it seems (not entirely sure if this should work properly, but let's see how far insights take us).
As you saw in my comment, $x$ can either be $9k-1$ or an integer root of $x^2 - x + 1 - 9k = 0$. Let's take the first case.
For $x = 9k -1$, we see $9 \mid x^3 + 1$, so turning our focus onto $(x+a)(x+b)(x+c)$, on putting $x = 9k -1$, we get $(9k - 1 + a)(9k - 1 +b)(9k - 1 + c)$
From what I can see of this, if any of $a,b,c$ is $1\pmod{9}$, then surely we $9$ divides the expression. But also look at the specific condition that has to be satisfied by $x$ : it has to be $-1 \pmod{9}$ and not all integers are $-1 \pmod{9}$.
I guess that you can perhaps proceed...
Note: before downvotes, please hint out to me my errors, because I am still in the basics of modular arithmetic.
