Given a + b + c = 14 and a^2 + b^2 + c^2 = 96, compute the exact value of (ab + bc + ca).

Introduction / Context:
This is a textbook application of the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca). Using the given sum and sum of squares, we can isolate the symmetric sum ab + bc + ca directly.

Given Data / Assumptions:

• a + b + c = 14.
• a^2 + b^2 + c^2 = 96.
• All quantities are real numbers.

Concept / Approach:
Square the total sum to link it with the sum of squares and twice the pairwise products. Solve for ab + bc + ca algebraically, keeping track of coefficients precisely.

Step-by-Step Solution:

Verification / Alternative check:
Plugging back into the identity reproduces 196 on the left, confirming correctness.

Why Other Options Are Wrong:
51, 55, 65, and 49 reflect arithmetic slips (wrong difference or a missing division by 2). Only 50 satisfies the identity exactly.

Common Pitfalls:
Forgetting the factor of 2 in front of (ab + bc + ca), or squaring 14 incorrectly.

Final Answer:
50