Question 157, Given that the five vertices of a regular quadrilateral pyramid P-ABCD lie on the same sphere. If the volume of the quadrilateral pyramid is 1, then the minimum value of the radius of the sphere is $\qquad$ _.
Question 157, Solution: As shown in the figure, let the radius of the circumscribed sphere of the regular quadrilateral pyramid P-ABCD be $r$, and the distance from the center of the sphere O to the base $\mathrm{ABCD}$ be $\mathrm{OM}=\mathrm{d}$. Then, $\mathrm{AM}=\sqrt{\mathrm{r}^{2}-\mathrm{d}^{2}}$, so the side length of the square $\mathrm{ABCD}$ is $\sqrt{2\left(\mathrm{r}^{2}-\mathrm{d}^{2}\right)}$. Therefore, the volume of the regular quadrilateral pyramid $P-A B C D$ is:
$$
\begin{array}{l}
\frac{1}{3} \cdot 2\left(r^{2}-d^{2}\right) \cdot(r+d)=1 \\
\Rightarrow 2(r-d) \cdot(r+d)^{2}=3 \\
\Rightarrow\left[\frac{2(r-d)+(r+d)+(r+d)}{3}\right]^{3} \geq 3 \\
\Rightarrow r \geq \frac{3 \sqrt[3]{3}}{4}
\end{array}
$$

Thus, the minimum value of the radius of the sphere is $\frac{3 \sqrt[3]{3}}{4}$.