Let x_1, x_2, ldots, x_6 be the roots of the | vedclass

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(C) The given equation is $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$.This is a geometric progression with $7$ terms.Multiplying by $(x-2)$,we get $(x-2)

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$|x_i|=2$ for all values of $i$

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(C) The given equation is $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$.This is a geometric progression with $7$ terms.Multiplying by $(x-2)$,we get $(x-2)(x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64) = 0$.This simplifies to $x^7 - 2^7 = 0$,which means $x^7 = 128$.The roots of this equation are $x_k = 2 e^{i \frac{2k\pi}{7}}$ for $k = 0, 1, 2, 3, 4, 5, 6$.However,the original equation is the sum of the geometric series excluding the term for $x=2$ (since $x=2$ makes the sum $64+64+64+64+64+64+64 = 448 
eq 0$).Thus,the roots are $x_k = 2 e^{i \frac{2k\pi}{7}}$ for $k = 1, 2, 3, 4, 5, 6$.For all these roots,$|x_k| = |2 e^{i \frac{2k\pi}{7}}| = 2 |e^{i \frac{2k\pi}{7}}| = 2(1) = 2$.Therefore,$|x_i|=2$ for all values of $i$.

Let $x_1, x_2, \ldots, x_6$ be the roots of the polynomial equation $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$. Then,

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