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If a + b + c = 8, a2 + b2 + c2 = 30, and a3 + b3 + c3 = 134, then the value of (abc)-1 is:
- A6
- B\(\frac{1}{{10}}\)
- C\(\frac{1}{6}\)
- D10
Answer not set
Discussion 1
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CCommunity
Given, a + b + c = 8, a2$ + b2 $+ c2 $= 30 and a3$ + b3 $+ c3 $= 134
? (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 82 = 30 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 64 - 30
⇒ (ab + bc + ca) = 34/2 = 17
We know that,
? a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – (ab + bc + ca))
⇒ 134 – 3abc = 8 × (30 – 17) = 104
⇒ 3abc = 134 - 104 = 30
⇒ abc = 30/3 = 10
∴ (abc)-1 = 1/10
? (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
⇒ 82 = 30 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 64 - 30
⇒ (ab + bc + ca) = 34/2 = 17
We know that,
? a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – (ab + bc + ca))
⇒ 134 – 3abc = 8 × (30 – 17) = 104
⇒ 3abc = 134 - 104 = 30
⇒ abc = 30/3 = 10
∴ (abc)-1 = 1/10