Question
Download Solution PDFఒకవేళ (a + b + c) = 20, మరియు aa2 + b2 + c2 = 152 అయితే, a3 + b3 + c3 - 3abc విలువను కనుగొనండి?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఉపయోగించిన సూత్రం:
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a3+b3+c3−3abc = (a + b + c) [a2 + b2 +c2 − (ab + bc + ca)]
గణన :
ఇక్కడ, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 400 = 152 + 2(ab + bc + ca)
⇒ 248 = 2(ab + bc+ ca)
⇒ (ab + bc + ca) = 124
ఇప్పుడు, a3 + b3 + c3 −3abc = (a + b + c) [a2 + b2 +c2 − (ab + bc + ca)]
⇒ 20 [152 - 124]
⇒ 28 × 20
⇒ 560
∴ సరైన సమాధానం 560.
Last updated on Jul 9, 2025
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