58 The top and bottom faces of a regular quadrilateral frustum have side lengths of positive integers $a$ and $b$ ($a<b$), and the height is 3. The lateral area is equal to the sum of the areas of the top and bottom faces. Then the volume of this regular quadrilateral frustum is
58 208. Let the height of the square frustum be $h$. From the problem, we have $4 \sqrt{\left(\frac{b-a}{2}\right)^{2}+h^{2}} \cdot \frac{a+b}{2}=a^{2}+b^{2}$, solving for $h$ gives $h=\frac{a b}{a+b}$. Given $h=3$, we get $a b-3 a-3 b+9=9$, which simplifies to $(a-3)(b-3)=9$. Since $a-3<b-3$, we have $a-3=1, b-3=9$. Therefore, $a=4, b=12$. It is easy to find that $V=208$.