If (a3 + b3 + c3 - 3abc) = 405, and (a + b + c) = 15, find the value of (a - b)2 + (b - c)2+(c - a)2

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SSC CGL 2023 Tier-I Official Paper (Held On: 20 Jul 2023 Shift 1)
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  1. 27
  2. 54
  3. 18
  4. 45

Answer (Detailed Solution Below)

Option 2 : 54
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Detailed Solution

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Given:

(a3 + b3 + c3 - 3abc) = 405, and (a + b + c) = 15

Concept used:

(a3 + b3 + c3 - 3abc) = 1/2 (a + b + c) [(a - b)2 + (b - c)2+(c - a)2]

Calculation:

(a3 + b3 + c3 - 3abc) = 1/2 (a + b + c) [(a - b)2 + (b - c)2+(c - a)2]

⇒ 405 = 1/2 × 15  (a - b)2 + (b - c)2+(c - a)2

⇒  (a - b)2 + (b - c)2+(c - a)2. = 810 / 15

⇒  (a - b)2 + (b - c)2+(c - a)2 = 54

∴ The correct option is 2

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