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If x^6=2 x^3-1 \text { and } x is not real then \sum_{r=1}^{50}\left(x^r+x^{2 r}\right)^3=

 

Option: 1

100


Option: 2

256


Option: 3

76


Option: 4

94


Answers (1)

best_answer

Given x6=2x3-1

x6-2x3+1=0

(x3-1)2=0

x3-1=0

x^3=1 \Rightarrow x=\omega, \omega^2 \quad x^r+x^{2 r}=\left\{\begin{array}{l} 2 \text { if } r \text { is a multiple of } 3 \\ -1 \text { if } r \text { is not a multiple of } 3 \end{array}\right.

On solving we get 94

 

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avinash.dongre

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