Let $f: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ be a function such that for all positive integers $m$ and $n$ with $m \ge n$, the following holds:
$$f(m \varphi(n^3)) = f(m) \cdot \varphi(n^3),$$
where $\varphi(n)$ denotes the Euler's totient function, which counts the number of positive integers up to $n$ that are coprime to $n$.

We start by examining the implications of the given functional equation for specific values of $n$.

1. **Case $n = 2$:**
$$f(4m) = 4f(m).$$

2. **Case $n = 2^2 = 4$:**
$$f(32m) = 32f(m).$$
Since $f(32m) = 4f(8m) = 16f(2m) = 32f(m)$, it follows that $f(2m) = 2f(m)$.

3. **Case $n = 3$:**
$$f(18m) = 18f(m) \implies f(9m) = 9f(m).$$

4. **Case $n = 3^2 = 9$:**
$$f(m \cdot 2 \cdot 3^5) = 2 \cdot 3^5 \cdot f(m) \implies f(m \cdot 3^5) = 3^5 \cdot f(m).$$
Since $f(9m) = 9f(m)$, it follows that $f(3m) = 3f(m)$.

By induction, we can generalize that for any prime $p$ and positive integer $m$:
$$f(mp^k) = p^k f(m).$$

Using this pattern, we assume $f(m) = km$ for some constant $k \in \mathbb{Z}^+$. We verify this by substituting back into the original functional equation:
$$f(m \varphi(n^3)) = k(m \varphi(n^3)) = k m \varphi(n^3) = f(m) \cdot \varphi(n^3).$$

Thus, the function $f$ that satisfies the given condition is:
$$f(m) = km \text{ for any positive integer constant } k.$$

The answer is: $\boxed{f(m)}$ = km $\text{}$ for any positive integer constant } k.}