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Question Number 47986 by behi83417@gmail.com last updated on 17/Nov/18

Commented by maxmathsup by imad last updated on 18/Nov/18

$2) (e) \Leftrightarrow \frac{1 - x^{8}}{1 - x} = 0 \Leftrightarrow x^{8} = 1 a n d x \neq 1 x = r e^{i θ} \Rightarrow r^{8} e^{i 8 θ} = e^{i 2 k π} \Rightarrow$ $r = 1 a n d 8 θ = 2 k π \Rightarrow x_{k} = e^{i \frac{k π}{4}} w i t h k \in [[1, 7]] a r e t h e r o o t s o f t h i s e q u a t i o n .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18

	$1) x^{4} + x^{3} + x^{2} + x + 1 = 0$ $x^{2} + x + 1 + \frac{1}{x} + \frac{1}{x^{2}} = 0$ $(x^{2} + \frac{1}{x^{2}}) + (x + \frac{1}{x}) + 1 = 0$ $y = x + \frac{1}{x}$ $y^{2} - 2 + y + 1 = 0$ $y^{2} + y - 1 = 0$ $y = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2} = \frac{- 1 + \sqrt{5}}{2} a n d \frac{- 1 - \sqrt{5}}{2}$ $s o x + \frac{1}{x} = y_{1} a n d x + \frac{1}{x} = y_{2}$ $y_{1} = \frac{- 1 + \sqrt{5}}{2} a n d y_{2} = \frac{- 1 - \sqrt{5}}{2}$ $x + \frac{1}{x} = y_{1}$ $x^{2} + 1 = {x y}_{1}$ $x^{2} - {x y}_{1} + 1 = 0$ $x = \frac{y_{1} \pm \sqrt{y_{1}^{2} - 4}}{2} = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{1 - 2 \sqrt{5 + 5}}{4} - 4}}{2}$ $x = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{1 - 2 \sqrt{5 + 5 - 16}}{4}}}{2}$ $x = \frac{\frac{- 1 + \sqrt{5}}{2} \pm \sqrt{\frac{- 10 - 2 \sqrt{5}}{4}}}{2}$ $x = \frac{- 1 + \sqrt{5} \pm i \sqrt{10 + 2 \sqrt{5}}}{4}$ $w h e n x + \frac{1}{x} = y_{2}$ $x^{2} - {x y}_{2} + 1 = 0$ $x = \frac{y_{2} \pm \sqrt{y_{2}^{2} - 4}}{2}$ $x = \frac{\frac{- 1 - \sqrt{5}}{2} \pm \sqrt{\frac{1 + 2 \sqrt{5 + 5}}{4} - 4}}{2}$ $= \frac{\frac{- 1 - \sqrt{5}}{2} \pm \sqrt{\frac{- 10 + 2 \sqrt{5}}{4}}}{2}$ $= \frac{- 1 - \sqrt{5} \pm i \sqrt{10 - 2 \sqrt{5}}}{4}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18

	$o r s o l u t i o n$ $x^{4} + x^{3} + x^{2} + x + 1 = 0$ $(x - 1) (x^{4} + x^{3} + x^{2} + x + 1) = 0$ $x^{5} - 1 = 0$ $x^{5} = 1 = c o s 2 k π + i s i n 2 π$ $x = c o s (\frac{2 k π}{5}) + i s i n (\frac{2 k π}{5})$ $s o r o o t s a r e . . . p u t k = 0, \pm 1, \pm 2$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Nov/18
	$2)x^6 (x+1)+x^4 (x+1)+x^2 (x+1)+1(x+1)=0 (x+1)(x^6 +x^4 +x^2 +1)=0 (x+1){x^4 (x^2 +1)+1(x^2 +1)}=0 (x+1)(x^2 +1)(x^4 +1)=0 when x+1=0 x=−1 x^2 +1=0 x^2 =−1=i^2 x=±i x^4 +1=0 x^4 =−1=cos(2kπ+π)+isin(2kπ+π) x=cos(((2k+1)/4)π)+isin(((2k+1)/4)π)$
	$2) x^{6} (x + 1) + x^{4} (x + 1) + x^{2} (x + 1) + 1 (x + 1) = 0$ $(x + 1) (x^{6} + x^{4} + x^{2} + 1) = 0$ $(x + 1) {x^{4} (x^{2} + 1) + 1 (x^{2} + 1)} = 0$ $(x + 1) (x^{2} + 1) (x^{4} + 1) = 0$ $w h e n x + 1 = 0$ $x = - 1$ $x^{2} + 1 = 0$ $x^{2} = - 1 = i^{2}$ $x = \pm i$ $x^{4} + 1 = 0$ $x^{4} = - 1 = c o s (2 k π + π) + i s i n (2 k π + π)$ $x = c o s (\frac{2 k + 1}{4} π) + i s i n (\frac{2 k + 1}{4} π)$
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