Regarding $x^3 -3xy+y^3=2005$ in integer solution

Question

I was reading about using modular arithmetic for Diophantine equations, and I stumble upon this problem:

Solve in integers the equation $x^3-3xy+y^3=2005$

The author takes the equation mod 9, so for the left side we have $2005 \equiv -2 (\mod9)$. Then he proceeds to state that for $3\mid xy$ then the left side would be $\equiv 0$ or $\equiv -1 , 3,$ or $4$ $(\mod9)$, and therefore for both cases (I don't know which both cases the author is talking about) the equation is impossible for modulo 9. However I don't understand what is the motivation behind using $3\mid xy$. (I'm aware that if we use that then $3xy$ would be $9k$ and that eliminates a term, but what if we assume $3\nmid$xy ?) And since $a^3 \equiv -1,0,1 (\mod 9)$. Shouldn't the remainders of $x^3+y^3$ be $ -1,0$ or $7$?

Answer 1

Based on the posted question, I surmise that the official solution is poorly written.

[E-1]:
Given any integer $~r,~$ you have that the $~\pmod{9}~$ congruence class of $~r^3~$ will either be $~-1, ~0,~$ or $~+1.~$

There are two possibilities: either $~9~$ divides $~3xy~$ or $~9~$ does not divide $~3xy.$

• Case 1 : $~9~$ divides $~3xy.$
  Then, you need two things to happen:
  You have to have that at least one of the two variables $~x~$ or $~y~$ is divisible by $~3.~$
  
  Simultaneously, you need $~x^2 + y^2 \equiv -2 \pmod{9}.$
  Based on [E-1], this is impossible, because you can't have (for example) $~x^3 \equiv -2 \pmod{9}.$

• Case 2 : $~9~$ does not divide $~3xy.$
  This implies that either $~-3xy \equiv 3\pmod{9}~$ or $~-3xy \equiv 6\pmod{9}.~$
  
  If $~-3xy \equiv 3\pmod{9},~$ then you need $~x^3 + y^3 \equiv 4 \pmod{9}.~$ Based on [E-1], this is impossible.
  
  If $~-3xy \equiv 6\pmod{9},~$ then you need $~x^3 + y^3 \equiv 1 \pmod{9}.~$
  
  Although [E-1] suggests that this is possible, it is in fact impossible, in this case, for the following reason. The only way to have $~x^3 + y^3 \equiv 1 \pmod{9},~$ is if exactly one of the two variables $~x,~$ or $~y,~$ is a multiple of $~3.~$ However, such a requirement contradicts the assumption of this bullet point, that $~9~$ does not divide $~3xy.$