Three positive numbers with pairwise products Given ab = 42, bc = 56, and ca = 48 for positive numbers a, b, c, find the value of the third number c.

Introduction / Context: Using pairwise products is a common technique to recover individual values of positive numbers. Here, three equations relate the three unknowns through products; careful algebra isolates one variable directly without solving a full system.

Given Data / Assumptions:

• ab = 42
• bc = 56
• ca = 48
• a, b, c are positive real numbers.

Concept / Approach: Multiply all three equations to get (ab)(bc)(ca) = (abc)^2. Then use a ratio to isolate c. A simpler method is to divide the product of the two equations involving c by the equation not involving c alone: c^2 = (bc * ca) / (ab).

Step-by-Step Solution:Compute c^2 = (bc * ca) / (ab).Plug values: c^2 = (56 * 48) / 42.Simplify: 56/42 = 4/3, so c^2 = (4/3) * 48 = 64.Therefore c = sqrt(64) = 8 (positive branch since numbers are positive).

Verification / Alternative check: If c = 8, then from bc = 56 ⇒ b = 7; from ca = 48 ⇒ a = 6. Check ab = 6 * 7 = 42, which matches. All conditions are satisfied.

Why Other Options Are Wrong:12, 6, 7: Substituting any of these values breaks at least one of the provided product equations when solving for the remaining variables.

Common Pitfalls: Adding products instead of multiplying/dividing; forgetting to take the positive square root; or arithmetic slips while simplifying 56*48/42. Keep factorization simple to avoid errors.

Final Answer: 8