If (a + b + c) = 14, and (a3 + b3 + c3 - 3abc) = 98, find the value of (a2 + b2 + c2).
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Correct answer: A
Explanation
The correct answer is option 1 i.e. 70.
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
Substitute a + b + c = 14 and a3 + b3 + c3 − 3abc = 98
⇒ 98 = 14 × (a2 + b2 + c2 − ab − bc − ca)
Divide both sides by 14:
⇒ 7 = a2 + b2 + c2 − ab − bc − ca
Use (a2 + b2 + c2 = (a + b + c)2 − 2(ab + bc + ca))
⇒ 7 = (a + b + c)2 − 2(ab + bc + ca) − ab − bc − ca
⇒ 7 = (142 - 2(ab + bc + ca)) − ab − bc − ca
⇒ 7 = (196 - 2(ab + bc + ca)) − ab − bc − ca
⇒ 7 = 196 − 3(ab + bc + ca)
⇒ 3(ab + bc + ca) = 189
⇒ ab + bc + ca = 189/3 = 63
So¸ ab + bc + ca = 63, substitute it back into the equation for a2 + b2 + c2 :
⇒ a2 + b2 + c2 = 196 − 2 × 63
⇒ a2 + b2 + c2 = 196 − 126 = 70