A) \[{{x}^{2}}-x-1=0\]
B) \[{{x}^{2}}-x+1=0\]
C) \[{{x}^{2}}+x-1=0\]
D) \[{{x}^{2}}+x+1=0\]
Correct Answer: D
Solution :
The roots of the equation x2 + x + 1 are given as \[\omega \And {{\omega }^{2}}.\]i,e. say, \[\alpha =\omega \And \beta ={{\omega }^{2}}\] \[{{\alpha }^{19}}={{\omega }^{19}}={{({{\omega }^{3}})}^{6}}\omega =\omega ;{{B}^{7}}=\] \[{{({{\omega }^{2}})}^{7=}}{{\omega }^{14}}={{({{\omega }^{3}})}^{4}}{{\omega }^{2}}={{\omega }^{2}}\] Hence the equation is \[{{x}^{2}}+x+1=0\]
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