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Let $S$ be the surface area and $V$ be the volume of the spherical balloon. Let $r$ be the radius of the balloon at time $t$. We are given that $\frac{dS}{dt} = 5 \text{ mm}^2/\text{s}$ and we need to find $\frac{dV}{dt}$ when $r = 8 \text{ mm}$.
The surface area of a sphere is given by $S = 4\pi r^2$ and the volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
Differentiating $S = 4\pi r^2$ with respect to $t$, we get: $$ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} $$
We are given $\frac{dS}{dt} = 5$ and $r = 8$. Substituting these values into the equation from Step 3, we get: $$ 5 = 8\pi (8) \frac{dr}{dt} $$ $$ \frac{dr}{dt} = \frac{5}{64\pi} $$
Differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$, we get: $$ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $$
Substituting $r = 8$ and $\frac{dr}{dt} = \frac{5}{64\pi}$ into the equation from Step 5, we get: $$ \frac{dV}{dt} = 4\pi (8^2) \left(\frac{5}{64\pi}\right) $$ $$ \frac{dV}{dt} = 4\pi (64) \left(\frac{5}{64\pi}\right) $$ $$ \frac{dV}{dt} = 20 $$
Final Answer: 20 mm³/s