$\exists X, Y, Z \in \mathbb{Z}[\omega]$ such that $X^3 + Y^3 + Z^3 = \omega$?

Question

I am considering the following problem: Denote by $\mathbb{Z}[\omega]$ the set of Eisenstein integers. Let $X, Y, Z \in \mathbb{Z}[\omega]$ be non-zero integers coprime to $1-\omega$. Is it possible to have $X^3 + Y^3 + Z^3 = \omega$? In attempting to answer this question, I have made note of the following:

1. $1-\omega$ is a prime number in the Eisenstein integers. (Does this then mean that $X, Y, Z$ are any number except for $1-\omega$ and the Eisenstein units $\pm 1, \pm \omega, \pm \omega^2$)?
2. In drawing the fundamental parallelogram associated to $(1-\omega)^2$, the Eisenstein integers inside are $0, 1, 2, 1+\omega, 2+\omega, 2+2\omega, -\omega, 1-\omega, -2\omega$. In drawing the addition and multiplication tables of these Eisenstein integers modulo $(1-\omega)^2$, it can be shown that the Eisenstein integers modulo $(1-\omega)^2$ are multiplicatively, but not additionally, "the same" as the rational integers modulo $9$.
3. In $\mathbb{Z}$, the cubic residues modulo $9$ are the invertible classes $[1], [8]$.

I am told that the problem should follow easily from my above work, but I cannot see how.

Answer 1

With $X=a_1+a_2\omega$, $Y=b_1+b_2\omega$, $Z=c_1+c_2\omega$, the equation $$ X^3+Y^3+Z^3-\omega=0 $$ is equivalent to $r+s\omega=0$ with $$ s=3a_1^2a_2 - 3a_1a_2^2 + 3b_1^2b_2 - 3b_1b_2^2 + 3c_1^2c_2 - 3c_1c_2^2 - 1 $$ Since we have $r=s=0$, but obviously $s\equiv 1 \bmod 3$, this is a contradiction.