<s>[INST] Show that if $n \equiv 4\pmod 9$, then $n$ cannot be written as the sum of three cubes. This is a repeat of a question asked here.

Show that if $n \equiv 4 (\mod 9)$, then $n$ cannot be written as the
  sum of three  cubes.

Solution: Any integer has least residue as either $0,1,2,3,4,5,6,7,8 (\mod 9)$
Now, $$0^3\equiv0 \pmod 9\\
1^3\equiv1 \pmod 9\\
2^3\equiv8 \pmod 9\\
3^3\equiv0 \pmod 9\\
4^3\equiv 1\pmod 9\\
5^3\equiv 8\pmod 9\\
6^3\equiv 0\pmod 9\\
7^3\equiv 1\pmod 9\\
8^3\equiv 8\pmod 9$$.
So,any integer cube is congruent to either $0,1 \text{or} 8 \pmod 9$
It is not possible to produce $4$ with the combination of $0,1,8$.
Closest we can get is $1+1+1=3$ or $0+0+8=8$.
My question (1) is this method correct?
Question (2) is: Why the answer to the linked question says that $m^3 \equiv 0, \pm1 \pmod9$?  How the user got "$-1$" and why "$8$" is missing?
 [/INST] HINT.- You have $8\equiv-1\pmod9$. Besides the same property goes for $n\equiv5\pmod9$, I mean
If $n\equiv5\pmod9$ then $n$ cannot be written as the sum of three cubes.
The proof is equal to the same as you have made for $n\equiv4\pmod9$
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