Two soap bubbles of radii 3 cm and 6 cm coalesce in a vacuum under isothermal conditions. What is the radius of the resulting bubble?

Let $R_1$ be the radius of the first bubble and $R_2$ be the radius of the second bubble.
$R_1 = 3 \text{ cm}$
$R_2 = 6 \text{ cm}$

For a soap bubble, the excess pressure inside is given by $\Delta P = \frac{4T}{R}$, where $T$ is the surface tension and $R$ is the radius. Since the bubbles are in a vacuum, the external pressure is 0, so the absolute pressure inside the bubble is $P = \frac{4T}{R}$.
The volume of a spherical bubble is $V = \frac{4}{3}\pi R^3$.
When two bubbles coalesce under isothermal conditions, the total number of moles of gas inside is conserved. From the ideal gas law ($PV = nRT$), since $T$ is constant, $n \propto PV$. Therefore, the product $PV$ is conserved:

where $P_f$ and $V_f$ are the pressure and volume of the final bubble.

Cancel the common factor $\left(\frac{4T \cdot 4\pi}{3}\right)$ from all terms:

Substitute the given values for $R_1$ and $R_2$:

The numerical value is approximately:

The radius of the bubble formed is \boxed{\text{3\sqrt{5} cm}}.