Actually, it is similar to solving linear congruences in $\mathbb Z/(n)$. In this case, you are starting with the principal ideal domain (PID) $\mathbb Z$, and expressing $Z/(n)$ as the sum of the prime-power factors, as $\mathbb Z/(72)\cong\mathbb Z/(8)\,\oplus\,\mathbb Z/(9)$, that is, you find solutions good modulo $8$ and $9$ separately and use Chinese Remainder Theorem. Right?
In your case I’ll use $\mathbb F_5$ to denote the field with five elements. It’s the same for the PID $\mathbb F_5[x]$ as it was for $\mathbb Z$. You have an element of the ring, namely $x^3-1$, that factors as $(x-1)(x^2+x+1)$, and you’ll need to solve your linear congruence modulo $x-1$ and modulo $x^2+x+1$. It’s messier, I’m sure, but the principle is exactly the same as in the familiar case.