Counting $3$-cycles in the Cartesian product of wheel and star graphs

Question

In getting $3$-cycles of the Cartesian product of two graphs, $\frac{1}{6} T(A)^3$ holds where $A$ is the adjacency matrix. On the other side, for star ($S_n, n>1$) and wheel graphs ($W_m, m>4$) there are exactly $n(m-1)$ $3$-cycles where $n$ and $m$ are the number of vertices of the $2$ graphs. Can we equate $\frac{1}{6}T(S_n \operatorname{\square} W_m)^3$ and $n(m-1)$, just to prove that the formula is true?