Question
Download Solution PDFIf (a + b + c) = 14, and (a3 + b3 + c3 - 3abc) = 98, find the value of (a2 + b2 + c2).
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
a + b + c = 14
a3 + b3 + c3 - 3abc = 98
Concept used:
a3 + b3 + c3 - 3abc = (a + b + c)[(a + b + c)2 - 3(ab + bc + ca)]
Calculation:
a3 + b3 + c3 - 3abc = (a + b + c)[(a + b + c)2 - 3(ab + bc + ca)]
⇒ 98 = 14 × [(14)2 - 3(ab + bc + ca)]
⇒ 7 = 196 - 3(ab + bc + ca)
⇒ 189 = 3(ab + bc + ca)
⇒ ab + bc + ca = 63
(a + b + c)2 = (a2 + b2 + c2) + 2(ab + bc + ca)]
(a2 + b 2 + c2) = 142 - 2 × 63
= 196 - 126
= 70
∴ The value of given identities is 70.
Last updated on Jul 23, 2025
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