Question
Download Solution PDFIf (a + b + c) = 20, and a2 + b2 + c2 = 152, find the value of a3 + b3 + c3 - 3abc.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFFormula Used :
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a3+b3+c3−3abc = (a + b + c) [a2 + b2 +c2 − (ab + bc + ca)]
Calculation :
Here, (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
Now, a3 + b3 + c3 −3abc = (a + b + c) [a2 + b2 +c2 − (ab + bc + ca)]
⇒ 20 [152 - 124]
⇒ 28 × 20
⇒ 560
∴ The correct answer is 560.
Last updated on Jul 2, 2025
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