10.  The positive integer pairs $(a, b)=$ $\qquad$ , where $a \neq b$, that make $\frac{a b^{2}}{a+b}$ a prime number.
Let $(a, b)=d \Rightarrow a=a_{1} d, b=b_{1} d$, substituting into $\frac{a b^{2}}{a+b}=p \Rightarrow a_{1} b_{1}^{2} d^{2}=p\left(a_{1}+b_{1}\right)$. Clearly, here we can only take $p=a_{1}, b_{1}=1, a_{1}+b_{1}=d^{2} \Rightarrow p=(d+1)(d-1) \Rightarrow d=2$. Therefore, $a_{1}=3 \Rightarrow a=6, b=2 \Rightarrow(a, b)=(6,2)$.