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?