10.  Make $\frac{a b^{2}}{a+b}(a \neq b)$ a prime number, the positive integer pair $(a, b)=$ $\qquad$ .
10. $(6,2)$.

Let $d=(a, b)$, $a=d u, b=d v\left(u 、 v \in \mathbf{Z}_{+},(u, v)=1\right)$.
Let $p=\frac{a b^{2}}{a+b}$. Then $d^{2} u v^{2}=p(u+v)$.
Since $(u+v, u)=(u+v, v)=(u, v)=1$,
we have $u v^{2} \mid p$.
Also, since $p$ is a prime, then $v=1$, and $u=1$ or $p$.
If $u=1$, then $d^{2}=2 p$.
Since $p$ is a prime, we have
$p\left|d \Rightarrow p^{2}\right| d^{2}$.
Thus, $p=2$, in this case, $(a, b)=(2,2)$ does not meet the requirement.
If $u=p$, then
$$
\begin{array}{l}
d^{2}=p+1 \Rightarrow(d-1)(d+1)=p \\
\Rightarrow\left\{\begin{array} { l } 
{ d - 1 = 1 , } \\
{ d + 1 = p }
\end{array} \Rightarrow \left\{\begin{array}{l}
d=2, \\
p=3 .
\end{array}\right.\right.
\end{array}
$$

Thus, $(a, b)=(6,2)$.
In conclusion, $(a, b)=(6,2)$.