Three. (This question is worth 50 points)
Find all pairs of positive integers $(a, b)(a>b)$ that satisfy
$$
(a-b)^{a b}=a^{b} \cdot b^{a}
$$
Three, Solution: Let $(a, b)=d, a=d p, b=d q, (p, q)=1, p, q \in N, p>q$, then
$$
(d(p-q))^{d^{2} p q}=(d p)^{d q} \cdot(d q)^{d p} \Rightarrow d^{d p q} \cdot(p-q)^{d p q}=d^{p+q} \cdot p^{q} \cdot q^{p}
$$

First, $d p q \geq p+q$ (otherwise, $d=q=1$ leads to a contradiction), thus $d^{d p q-p-q} \cdot(p-q)^{d p q}=p^{q} \cdot q^{p}$, so $(p-q) \mid p^{q} \cdot q^{p}$. Since $(p-q, p)=(p-q, q)=1$, it must be that $p-q=1$. Substituting into the original equation, we have $d^{d p q-p-q}=p^{q} \cdot q^{q+1}$. Since $(p, q)=1$, we can set $d=m n, m, n \in N_{+}$, where $m, p$ have the same prime factors, and $n, q$ have the same prime factors, thus $(m, n)=1$. This gives us $m^{d p q-p-q}=p^{q}=(q+1)^{q}$. Since $(d p q-p-q, q)=1$, we can set $m=t^{q}, t \in N$, substituting in we get $t^{d p q-p-q}=q+1>1$, so $t \geq 2$.
If $q=2$, then $t^{6 d-5}=3$, giving $t=3, d=1$, thus $q=2, p=3$, which is discarded!
If $q \geq 3$, then $d p q-p-q=d q(q+1)-(2 q+1) \geq 3(q+1)-(2 q+1)=q+2$, so $q+1=t^{d p q-p-q} \geq t^{q+2} \geq 2^{q+2} \geq q+2$, a contradiction!
Therefore, it must be that $q=1, p=2$, so $t^{2 d-3}=2$, giving $t=2, d=2$, at this point $(a, b)=(4,2)$. In summary, the original equation has a unique solution $(a, b)=(4,2)$.