A solid metallic sphere of radius 12 cm is melted and recast into three smaller solid spheres. If two of the new spheres have radii 6 cm and 8 cm respectively, what is the radius (in cm) of the third sphere?

Introduction / Context:
When a solid is melted and recast, total volume is conserved (no loss of material). Here, one large sphere is melted into three smaller spheres. Using the sphere volume formula, we equate the original volume to the sum of the three new volumes and solve for the unknown radius.

Given Data / Assumptions:

• Original radius R = 12 cm.
• Two new radii r1 = 6 cm and r2 = 8 cm.
• Third radius r3 = ?
• Use volume of sphere: V = (4/3) * π * r^3.

Concept / Approach:
Volume conservation: (4/3)πR^3 = (4/3)πr1^3 + (4/3)πr2^3 + (4/3)πr3^3 ⇒ R^3 = r1^3 + r2^3 + r3^3. Cancel (4/3)π from both sides.

Step-by-Step Solution:

Verification / Alternative check:
Compute exact volumes: (4/3)π(1728) on LHS and (4/3)π(216 + 512 + 1000) on RHS. The sums match exactly, confirming r3 = 10 cm.

Why Other Options Are Wrong:

• 12 cm: Would imply zero melting into smaller spheres; contradicts given data.
• 14 cm or 16 cm: Their cubes greatly exceed the remaining volume after subtracting the first two spheres.

Common Pitfalls:
Forgetting volume scales with the cube of radius; using surface area instead of volume; arithmetic slips when summing cubes.

Final Answer:
10 cm