If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165, then

1.	If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165, then the value of 4abc is:1. -12. -43. 14. 4
Answer» Correct Answer - Option 4 : 4 Given: If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165 Concept 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: ⇒ (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca) ⇒ 6² = 30 + 2 (ab + bc + ca) ⇒ (ab + bc + ca) = 6/2 = 3 ⇒ a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca) ⇒ 165 - 3abc = 6 × (30 - 3) ⇒ 165 - 3abc = 6 × 27 ⇒ 3abc = 165 - 162 ⇒ abc = 1 ∴ 4abc = 4 × 1 = 4

If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165, then the value of 4abc is:1. -12. -43. 14. 4

Answer» Correct Answer - Option 4 : 4

Given:

If a + b + c = 6, a2 + b2 + c2 = 30 and a3 + b3 + c3 = 165

Concept 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:

⇒ (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

⇒ 6² = 30 + 2 (ab + bc + ca)

⇒ (ab + bc + ca) = 6/2 = 3

⇒ a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

⇒ 165 - 3abc = 6 × (30 - 3)

⇒ 165 - 3abc = 6 × 27

⇒ 3abc = 165 - 162

⇒ abc = 1

∴ 4abc = 4 × 1 = 4

Discussion

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