Myhill-Nerode equivalence classes
To begin with, note the the Myhill-Nerode equivalence classes for a language $L$ is the same as the Myhill-Nerode classes for its complement $\overline{L}$. This follows immediately by definition, and I leave it you (it also appears here).
So given $\Sigma = \{a, b, c\}$, we need to find the Myhill-Nerode classes for $\overline{L} = \{ \epsilon, cc\}\cup \{cca^i: i\geq 2 \}$. There are many ways to do so, one can, for example, consider what happens in a minimal deterministic automaton for the language - it is known that the states of the minimal deterministic automaton correspond to the Myhill-Nerode classes. Yet, the given language is simple enough, you can actually experiment with it a bit and show that the following are the equivalence class for $\overline{L}$ (indeed, $\overline{L}$ has a simple structure - every word of length at least $4$ in the laguage is of the form $cca^i$, so clearly all these words are equivalent. Then, you can look at words of length &lt; 4, and try to see which among them are equivalent, etc.):

$C_1 = \{ \epsilon\}$.
$C_2 =  \{ c\}$.
$C_3 =  \{ cc\}$.
$C_4 = \{ cca\}$.
$C_5 = \{ cca^i: i\geq 2\}$.
$C_6 = \Sigma^*\setminus \left( \bigcup_{i\in [5]} C_i\right)$.

Clearly the 5 sets above partition $\Sigma^*$, so to show that they are indeed the Myhill-Nerode classes, we need to show that all the words in the same set are Myhill-Nerode equivalent, and for every $ 1\leq i &lt; j \leq 6$, it holds that the words in the set $C_i$ are not equivalent to the words in the set $C_j$.
There you go, try to prove that the above sets are indeed the equivalence classes. That should be easy enough now.