If (a + b + c) = 14 and (a3 + b3 + c3 − 3abc) = 98, then find the value of (ab…

Concept: For any three numbers, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx), and the sum-of-cubes identity x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx) links the sum of cubes minus thrice the product to the sum times the “sum of squares minus sum of pairwise products” factor.

1. Let x = a2 + b2 + c2 and y = ab + bc + ca. Squaring the given sum: (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), so 142 = 196 = x + 2y … (i)

2. Substitute a + b + c = 14 into the identity a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca): 98 = 14 × (x − y), so x − y = 98 / 14 = 7 … (ii)

3. Subtract (ii) from (i): (x + 2y) − (x − y) = 196 − 7, so 3y = 189, giving y = ab + bc + ca = 63.

Cross-check: using the alternate identity (a + b + c)3 = a3 + b3 + c3 + 3(a + b + c)(ab + bc + ca) − 3abc: 143 = 2744, and a3 + b3 + c3 − 3abc = 98, so 2744 = 98 + 3 × 14 × (ab + bc + ca), giving 3 × 14 × (ab + bc + ca) = 2646 and ab + bc + ca = 2646 / 42 = 63 — the same result confirms the value.

Hence, ab + bc + ca = 63.