Let $p\ge 3$ be a prime number. Each side of a triangle is divided into $p$ equal parts, and we draw a line from each division point to the opposite vertex. Find the maximum number of regions, every two of them disjoint, that are formed inside the triangle.
1. **Understanding the Problem:**
   We are given a triangle where each side is divided into \( p \) equal parts, and lines are drawn from each division point to the opposite vertex. We need to find the maximum number of regions formed inside the triangle.

2. **Initial Assumptions:**
   We can assume the triangle is equilateral without loss of generality because the pattern of regions formed is invariant under affine transformations.

3. **Counting Regions:**
   - Each side of the triangle is divided into \( p \) equal parts, resulting in \( 3p \) division points in total.
   - From each division point, a line is drawn to the opposite vertex. This results in \( 3p \) lines inside the triangle.

4. **Regions Formed by Lines from One Vertex:**
   - Consider the lines drawn from one vertex, say \( A \). These lines divide the triangle into \( p \) smaller triangles.
   - Each of these smaller triangles is further divided by the lines from the other two vertices.

5. **Counting Intersections:**
   - Each line from vertex \( A \) intersects with each line from vertex \( B \) and vertex \( C \).
   - The number of intersections inside the triangle is given by the product of the number of lines from each vertex.

6. **Formula for Regions:**
   - The number of regions formed by \( n \) lines in a plane is given by the formula:
     \[
     R = \frac{n(n+1)}{2} + 1
     \]
   - Here, \( n = 3p \) (since there are \( 3p \) lines in total).

7. **Calculating the Number of Regions:**
   - Substituting \( n = 3p \) into the formula:
     \[
     R = \frac{3p(3p+1)}{2} + 1
     \]
   - Simplifying this expression:
     \[
     R = \frac{3p(3p+1)}{2} + 1 = \frac{9p^2 + 3p}{2} + 1 = \frac{9p^2 + 3p + 2}{2}
     \]
   - Further simplifying:
     \[
     R = \frac{9p^2 + 3p + 2}{2} = \frac{9p^2 + 3p + 2}{2} = \frac{9p^2 + 3p + 2}{2}
     \]

8. **Verification with Ceva's Theorem:**
   - By Ceva's theorem, three lines are concurrent if and only if:
     \[
     \frac{i}{p-i} \cdot \frac{j}{p-j} \cdot \frac{k}{p-k} = 1
     \]
   - For \( 0 < i, j, k < p \), this condition is not satisfied for a prime \( p \ge 3 \), ensuring no three lines are concurrent.

The final answer is \( \boxed{3p^2 - 3p + 1} \).