Question
If a + b + c = 12 and ab + bc + ca = 47, and a, b, c are
real numbers, find the value of a³ + b³ + c³ − 3abc.Solution
Use identity: a³ + b³ + c³ − 3abc = (a + b + c)[a² + b² + c² − ab − bc − ca] First find a² + b² + c²: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) ⇒ 12² = a² + b² + c² + 2×47 144 = a² + b² + c² + 94 a² + b² + c² = 144 − 94 = 50 Then, a² + b² + c² − (ab + bc + ca) = 50 − 47 = 3 So, a³ + b³ + c³ − 3abc = (a + b + c) × 3 = 12 × 3 = 36.
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