Let radius of the 3rd spherical ball be R

Given Information:
- Diameter of the original ball = 3 cm

To find:
- Diameter of the third ball

Solution:
1. Volume Conservation:
- When a solid is melted and recast into smaller parts, the total volume remains the same.
- In this case, the volume of the original ball is equal to the sum of the volumes of the three smaller balls.
- Since the balls are spherical, we can use the formula for the volume of a sphere: V = (4/3)πr³, where V is the volume and r is the radius.

2. Volume of the Original Ball:
- The diameter of the original ball is 3 cm, so the radius is half of the diameter, which is 3/2 = 1.5 cm.
- Using the formula for the volume of a sphere, V = (4/3)π(1.5)³ = (4/3)π(3.375) = 14.25π cm³.

3. Volume of the Three Smaller Balls:
- Let the radius of the first smaller ball be r₁ cm.
- The diameter of the first smaller ball is 1.5 cm, so the radius is half of the diameter, which is 1.5/2 = 0.75 cm.
- Using the formula for the volume of a sphere, V₁ = (4/3)π(0.75)³ = (4/3)π(0.421875) = 1.40625π cm³.

- Let the radius of the second smaller ball be r₂ cm.
- The diameter of the second smaller ball is 2 cm, so the radius is half of the diameter, which is 2/2 = 1 cm.
- Using the formula for the volume of a sphere, V₂ = (4/3)π(1)³ = (4/3)π = 4.18879π cm³.

- Let the radius of the third smaller ball be r₃ cm.
- We need to find the diameter of the third ball, which is twice the radius.
- Let the diameter of the third ball be D cm.
- The radius of the third ball is half of the diameter, which is D/2 cm.
- Using the formula for the volume of a sphere, V₃ = (4/3)π(D/2)³ = (4/3)π(D³/8) = (1/6)πD³ cm³.

4. Equating V and V₁ + V₂ + V₃:
- V = V₁ + V₂ + V₃
- 14.25π = 1.40625π + 4.18879π + (1/6)πD³
- 14.25 = 1.40625 + 4.18879 + (1/6)D³
- 14.25 - 5.59504 = (1/6)D³
- 8.65496 = (1/6)D³
- D³ = 8.65496 * 6
- D³ = 51.92976
- D = ∛(51.92976)