If a, b, c, x, y, z are real numbers such that (a + b + c)2 - 3(ab + bc + ca) + 3(x2 + y2 + z2) = 0, then which one of the following is correct?

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CDS Elementary Mathematics 3 Sep 2023 Official Paper
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  1. a = b = c, x = y = z ≠ 0
  2. a = b = c = 0, x = y = z = 1
  3. a = b = c, x = y = z = 0
  4. a ≠ b ≠ c, x = y = z = 0

Answer (Detailed Solution Below)

Option 3 : a = b = c, x = y = z = 0
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Detailed Solution

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Given:

(a + b + c)- 3(ab + bc + ca) + 3(x+ y+ z2) = 0

Calculation:

We have

(a + b + c)- 3(ab + bc + ca) + 3(x+ y+ z2) = 0

We solve this question by options 

If we put the value of option 3 then

a = b = c  = k (let) and x = y = z = 0

(k + k + k)2 - 3(k2 + k2 + k2) + 3(0)

⇒ (3k)2 - 3(3k2

⇒ 9k2 - 9k2 = 0

∴ The correct option is 3.

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