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If a + b + c \(=2,\frac{1}{a} + \frac{1}{b}+\frac {1}{c}=0, ac=\frac{4}{b}\) and a3 + b3 + c3 = 28, find the value of a2 + b2 + c2
1. 8
2. 6
3. 10
4. 12

1 Answer

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Best answer
Correct Answer - Option 1 : 8

Given:

a + b + c = 2, 1/a + 1/b + 1/c = 0, ac = 4/b and a3 + b3 + c3 = 28

Formula:

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

Calculation:

1/a + 1/b + 1/c = 0

⇒ (ab + bc + ca)/abc = 0

⇒ ab + bc + ca = 0

Also,

ac = 4/b

⇒ abc = 4

According to the question

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

⇒ 28 - 3 × 4 = 2 × (a2 + b2 + c2 - 0)

⇒ (28 - 12)/2 = a2 + b2 + c2

⇒ a2 + b2 + c2 = 16/2

⇒ a2 + b2 + c2 = 8

∴ The value of a2 + b2 + c2 is 8. 

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