For a prime $p$ of the form $12k+1$ and $\mathbb{Z}_p=\{0,1,2,\cdots,p-1\}$, let 
\[\mathbb{E}_p=\{(a,b) \mid a,b \in \mathbb{Z}_p,\quad  p\nmid 4a^3+27b^2\}\]
For $(a,b), (a',b') \in \mathbb{E}_p$ we say that $(a,b)$ and $(a',b')$ are equivalent if there is a non zero element $c\in \mathbb{Z}_p$ such that $p\mid (a' -ac^4)$ and $p\mid (b'-bc^6)$. Find the maximal number of inequivalent elements in $\mathbb{E}_p$.
1. **Determine the size of $\mathbb{E}_p$:**

   We start by finding the number of pairs \((a, b)\) in \(\mathbb{Z}_p \times \mathbb{Z}_p\) such that \(4a^3 + 27b^2 \not\equiv 0 \pmod{p}\). There are \(p^2\) total pairs \((a, b)\) in \(\mathbb{Z}_p \times \mathbb{Z}_p\).

   We need to count the number of pairs \((a, b)\) such that \(4a^3 + 27b^2 \equiv 0 \pmod{p}\). This is equivalent to finding the number of solutions to the equation \(4a^3 + 27b^2 \equiv 0 \pmod{p}\).

   \[
   4a^3 + 27b^2 \equiv 0 \pmod{p} \implies 27b^2 \equiv -4a^3 \pmod{p} \implies b^2 \equiv -\frac{4}{27}a^3 \pmod{p}
   \]

   If \(a = 0\), then \(b = 0\) is the only solution. If \(a \neq 0\), we need to determine whether \(-\frac{4}{27}a^3\) is a quadratic residue modulo \(p\).

   Since \(p \equiv 1 \pmod{12}\), \(p\) can be written as \(p = 12k + 1\). This implies that \(p \equiv 1 \pmod{3}\) and \(p \equiv 1 \pmod{4}\). Therefore, \(-\frac{4}{27}a^3\) is a quadratic residue for \(\frac{p-1}{2}\) values of \(a\) and not a quadratic residue for the other \(\frac{p-1}{2}\) values of \(a\).

   For each \(a\) such that \(-\frac{4}{27}a^3\) is a quadratic residue, there are 2 corresponding values of \(b\). Therefore, there are \(2 \cdot \frac{p-1}{2} = p-1\) solutions for \(a \neq 0\).

   Adding the solution for \(a = 0\), we get \(p\) solutions in total. Thus, the number of pairs \((a, b)\) such that \(4a^3 + 27b^2 \equiv 0 \pmod{p}\) is \(p\).

   Therefore, the number of pairs \((a, b)\) such that \(4a^3 + 27b^2 \not\equiv 0 \pmod{p}\) is:

   \[
   |\mathbb{E}_p| = p^2 - p
   \]

2. **Determine the number of inequivalent elements in \(\mathbb{E}_p\):**

   We need to find the number of inequivalent elements in \(\mathbb{E}_p\). Two pairs \((a, b)\) and \((a', b')\) are equivalent if there exists a non-zero element \(c \in \mathbb{Z}_p\) such that:

   \[
   a' \equiv ac^4 \pmod{p} \quad \text{and} \quad b' \equiv bc^6 \pmod{p}
   \]

   We consider three cases:

   **Case (i): \(a = 0, b \neq 0\)**

   For \(a = 0\), \(b \neq 0\), the equivalence condition simplifies to \(b' \equiv bc^6 \pmod{p}\). Since \(c^6\) can take \(\frac{p-1}{6}\) distinct values in \(\mathbb{Z}_p\), there are:

   \[
   \frac{p-1}{\frac{p-1}{6}} = 6
   \]

   inequivalent elements.

   **Case (ii): \(a \neq 0, b = 0\)**

   For \(a \neq 0\), \(b = 0\), the equivalence condition simplifies to \(a' \equiv ac^4 \pmod{p}\). Since \(c^4\) can take \(\frac{p-1}{4}\) distinct values in \(\mathbb{Z}_p\), there are:

   \[
   \frac{p-1}{\frac{p-1}{4}} = 4
   \]

   inequivalent elements.

   **Case (iii): \(a \neq 0, b \neq 0\)**

   For \(a \neq 0, b \neq 0\), the equivalence condition involves both \(c^4\) and \(c^6\). Since \(c^2\) can take \(\frac{p-1}{2}\) distinct values in \(\mathbb{Z}_p\), there are:

   \[
   \frac{p^2 - p - 2(p-1)}{\frac{p-1}{2}} = 2p - 4
   \]

   inequivalent elements.

3. **Combine the results from the three cases:**

   Adding the number of inequivalent elements from all three cases, we get:

   \[
   6 + 4 + (2p - 4) = 2p + 6
   \]

   Therefore, the maximal number of inequivalent elements in \(\mathbb{E}_p\) is:

   \[
   \boxed{2p + 6}
   \]