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If a2 + b2 + c2 = 18% of 211(1/9) and ab + bc + ca = –(a2 + b2 + c2)/2, then find the value of a3 + b3 + c3.
1. abc/3
2. abc/2
3. 3abc
4. 2abc

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Correct Answer - Option 3 : 3abc

Given,

a2 + b2 + c2 = 18% of 211(1/9)

a2 + b2 + c2 = 18 × (1900/900) = 38

Also given,

ab + bc + ca = –(a2 + b2 + c2)/2

ab + bc + ca = –38/2 = –19

We know that:

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

(a + b + c)2 = 38 + 2(–19)

(a + b + c)2 = 0

a + b + c = 0

We also know that:

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

a3 + b3 + c3 – 3abc = 0

a3 + b3 + c3 = 3abc

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