Consider a simple undirected unweighted graph with at least three vertices. If…

Key idea: relate closed walks of length three to 3-cycles.

(A^3)_{ii} counts the number of walks of length three that start and end at vertex i.

1. Each triangle (3-cycle) with vertices u, v, w produces closed walks of length three.

2. For a given triangle, there are three choices of starting vertex (u, v, or w).

3. For each starting vertex there are two directions to traverse the triangle (clockwise and counterclockwise).

4. Therefore each triangle contributes 3 × 2 = 6 closed walks of length three, so it contributes 6 to the trace of A^3.

Summing over all vertices, trace(A^3) equals the total number of closed walks of length three, which is 6 times the number of distinct 3-cycles.

Conclusion: the number of 3-cycles = trace(A^3) / 6.

Note: This holds for a simple undirected graph without self-loops or multiple edges; the adjacency matrix is symmetric and has zeros on the diagonal.