$I f a = \cos \frac{2 π}{7} + i \sin \frac{2 π}{7}, α = a + a^{2} + a^{4} a n d β = a^{3} + a^{5} + a^{6} t h e n α, β a r e t h e r o o t s o f t h e e q u a t i o n$

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$x^{2} + x + 1 = 0$

$x^{2} + 2 x + 2 = 0$

$x^{2} + x + 2 = 0$

$x^{2} + 2 x + 3 = 0$

answer is B.

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Detailed Solution

$\begin{array}{l} I f a = \cos \frac{2 π}{7} + i \sin \frac{2 π}{7} t h e n \\ a i s r o o t o f x^{7} - 1 = 0 \Rightarrow a^{7} = 1 \\ α = a + a^{2} + a^{4}, β = a^{3} + a^{5} + a^{6} \\ α + β = a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} \\ = - 1 [1 + a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6}] = 0 \\ α β = (a + a^{2} + a^{4}) (a^{3} + a^{5} + a^{6}) \\ = a^{4} + a^{6} + a^{7} + a^{5} + a^{7} + a^{8} + a^{7} + a^{9} + a^{10} \\ = 3 (a^{7}) + a^{7} . a^{1} + a^{7} . a^{2} + a^{7} . a^{3} + a^{4} + a^{5} + a^{6} \\ = 3 + a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} \\ = 3 - 1 \\ = 2 \\ Re q u i r e d E q u a t i o n x^{2} - (α + β) x + α β = 0 \\ \Rightarrow x^{2} + x + 2 = 0 \end{array}$