A) 12
B)
C) 9
D) 14
Correct Answer: B
Solution :
Given equation is \[{{x}^{2}}+x+1=0.\] On solving equation, we get \[\text{x= }\!\!\omega\!\!\text{ and x=}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\] Case \[\text{I}\] When \[x=\omega \] Then \[\sum\limits_{n=1}^{6}{{{\left[ {{x}^{n}}+\frac{1}{{{x}^{n}}} \right]}^{2}}=\sum\limits_{n=1}^{6}{{{[{{\omega }^{n}}+{{\omega }^{2n}}]}^{2}}}\left[ \because \frac{1}{\omega }={{\omega }^{2}} \right]}\] \[={{(\omega +{{\omega }^{2}})}^{2}}+{{({{\omega }^{2}}+{{\omega }^{4}})}^{2}}+{{({{\omega }^{3}}+{{\omega }^{6}})}^{2}}\] \[+{{({{\omega }^{4}}+{{\omega }^{8}})}^{2}}+{{({{\omega }^{5}}+{{\omega }^{10}})}^{2}}+{{({{\omega }^{6}}+{{\omega }^{12}})}^{2}}\] \[={{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( 2 \right)}^{2}}=12\]Case \[\text{II}\] When \[x={{\omega }^{2}}\] Then \[\sum\limits_{n=1}^{6}{{{\left[ {{x}^{n}}+\frac{1}{{{x}^{n}}} \right]}^{2}}=\sum\limits_{n=1}^{6}{{{[{{\omega }^{2n}}+{{\omega }^{n}}]}^{2}}}}\] \[\left[ \because \frac{1}{{{\omega }^{2}}}=\omega \right]\] = 12
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