A sphere and a cone have the same volume. The area of the lateral surface of the cone is $80 \pi$ and the total surface area of the cone is $144 \pi$. Determine the radius of the sphere.

(The volume of a sphere with radius $r$ is $\frac{4}{3} \pi r^{3}$. The volume of a cone with radius $r$ and height $h$ is $\frac{1}{3} \pi r^{2} h$. The surface area of a cone consists of two parts: its base and its lateral surface. The lateral surface area of a cone with radius $r$ and slant height $s$ is $\pi r s$.)
Suppose that the cone has radius $r$, height $h$, and slant height $s$.

![](https://cdn.mathpix.com/cropped/2024_04_30_68642ad613c12af1e439g-09.jpg?height=442&width=311&top_left_y=1622&top_left_x=972)

Since the lateral surface area of the cone is $80 \pi$ and the total surface area is $144 \pi$, then the area of the base of the cone is $144 \pi-80 \pi=64 \pi$.

Thus, $\pi r^{2}=64 \pi$, which gives $r^{2}=64$ or $r=8$ (since $r>0$ ).

Since the lateral surface area is $80 \pi$ and $r=8$, then $\pi r s=80 \pi$ or $8 s=80$ which gives $s=10$.

Now, by the Pythagorean Theorem, $s^{2}=r^{2}+h^{2}$ and so $h^{2}=s^{2}-r^{2}=10^{2}-8^{2}=36$.

Since $h>0$, then $h=6$.

Thus, the volume of the cone is $\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi 8^{2}(6)=128 \pi$.

Suppose that a sphere of equal volume has radius $R$. Then $\frac{4}{3} \pi R^{3}=128 \pi$.

Thus, $R^{3}=\frac{3}{4} \cdot 128=96$, and so $R=\sqrt[3]{96}$.

ANSWER: $\sqrt[3]{96}$