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Question Number 52025 by Tawa1 last updated on 02/Jan/19

Commented by Abdo msup. last updated on 03/Jan/19
$8)(x+1)^6 +(x−1)^6 =0 ⇔(((x−1)^6 )/((x+1)^6 )) =−1 ⇔ (((x−1)/(x+1)))^6 =−1 let ((x−1)/(x+1)) =z ⇒z^6 =−1 ⇒ z^6 =e^(i(2k+1)π) ⇒the roots of this (e) are z_k =e^(i(((2k+1)π)/6)) and k∈[[0,5]] ⇒the roots of the first (e) are x_k / ((x_k −1)/(x_k +1)) =z_k ⇒x_k −1 =z_k x_k +z_k ⇒ (1−z_k )x_k = 1+z_k ⇒x_k =((1+z_k )/(1−z_k )) =((1+cos(((2k+1)/6)π) +isin((((2k+1)π)/6)))/(1−cos((((2k+1)π)/6))−isin((((2k+1)π)/6)))) =((2cos^2 ((((2k+1)π)/(12))) +2isin((((2k+1)π)/(12)))cos((((2k+1)π)/(12))))/(2sin^2 ((((2k+1)π)/(12)))−2isin((((2k+1)π)/(12)))cos((((2k+1)π)/(12))))) =(1/(tan((((2k+1)π)/(12))))) (e^(i(((2k+1)π)/(12))) /(−i e^((i(2k+1)π)/(13)) )) =icotan((((2k+1)π)/(12))) ⇒ the roots are x_k =i cotan((((2k+1)π)/(12))) with k∈{0,1,2,3,4,5}$
$8) {(x + 1)}^{6} + {(x - 1)}^{6} = 0 \Leftrightarrow \frac{{(x - 1)}^{6}}{{(x + 1)}^{6}} = - 1 \Leftrightarrow$ ${(\frac{x - 1}{x + 1})}^{6} = - 1 l e t \frac{x - 1}{x + 1} = z \Rightarrow z^{6} = - 1 \Rightarrow$ $z^{6} = e^{i (2 k + 1) π} \Rightarrow t h e r o o t s o f t h i s (e) a r e$ $z_{k} = e^{i \frac{(2 k + 1) π}{6}} a n d k \in [[0, 5]] \Rightarrow t h e r o o t s o f t h e f i r s t$ $(e) a r e x_{k} / \frac{x_{k} - 1}{x_{k} + 1} = z_{k} \Rightarrow x_{k} - 1 = z_{k} x_{k} + z_{k} \Rightarrow$ $(1 - z_{k}) x_{k} = 1 + z_{k} \Rightarrow x_{k} = \frac{1 + z_{k}}{1 - z_{k}}$ $= \frac{1 + c o s (\frac{2 k + 1}{6} π) + i s i n (\frac{(2 k + 1) π}{6})}{1 - c o s (\frac{(2 k + 1) π}{6}) - i s i n (\frac{(2 k + 1) π}{6})}$ $= \frac{2 {c o s}^{2} (\frac{(2 k + 1) π}{12}) + 2 i s i n (\frac{(2 k + 1) π}{12}) c o s (\frac{(2 k + 1) π}{12})}{2 {s i n}^{2} (\frac{(2 k + 1) π}{12}) - 2 i s i n (\frac{(2 k + 1) π}{12}) c o s (\frac{(2 k + 1) π}{12})}$ $= \frac{1}{t a n (\frac{(2 k + 1) π}{12})} \frac{e^{i \frac{(2 k + 1) π}{12}}}{- i e^{\frac{i (2 k + 1) π}{13}}} = i c o t a n (\frac{(2 k + 1) π}{12})$ $\Rightarrow t h e r o o t s a r e x_{k} = i c o t a n (\frac{(2 k + 1) π}{12}) w i t h$ $k \in {0, 1, 2, 3, 4, 5}$
Commented by Tawa1 last updated on 03/Jan/19

$God bless you sir$
Commented by Tawa1 last updated on 03/Jan/19

$Sir, what of the deduction . Thanks for your time$
Commented by maxmathsup by imad last updated on 03/Jan/19

$y o u a r e w e l c o m e s i r .$
Commented by maxmathsup by imad last updated on 03/Jan/19

$w e h a v e S = {t a n}^{2} (\frac{π}{12}) + {t a n}^{2} (\frac{3 π}{12}) + {t a n}^{2} (\frac{5 π}{12}) = {t a n}^{2} (\frac{π}{12}) + {t a n}^{2} (\frac{π}{2} - \frac{π}{12}) + 1$ $= {t a n}^{2} (\frac{π}{12}) + \frac{1}{{t a n}^{2} (\frac{π}{12})} + 1 b u t t a n (\frac{π}{12}) = 2 - \sqrt{3} \Rightarrow$ $S = {(2 - \sqrt{3})}^{2} + \frac{1}{{(2 - \sqrt{3})}^{2}} + 1 = {(2 - \sqrt{3})}^{2} + {(2 + \sqrt{3})}^{2} + 1 = 7 - 4 \sqrt{3} + 7 + 4 \sqrt{3} + 1$ $= 15 .$
Commented by maxmathsup by imad last updated on 03/Jan/19

${c o s}^{2} (\frac{π}{12}) = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} \Rightarrow c o s (\frac{π}{12}) = \frac{\sqrt{2 + \sqrt{3}}}{2}$ $s i n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2} \Rightarrow t a n (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} = \frac{\sqrt{2 - \sqrt{3}} \sqrt{2 - \sqrt{3}}}{\sqrt{2^{2} - 3}} = 2 - \sqrt{3} .$
Commented by Tawa1 last updated on 03/Jan/19

$God bless you sir . i appreciate$
Commented by Abdo msup. last updated on 13/Jan/19

$m o s t w e l c o m e .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

	$7) x^{9} - x^{5} + x^{4} - 1 = 0$ $x^{5} (x^{4} - 1) + 1 (x^{4} - 1) = 0$ $(x^{4} - 1) (x^{5} + 1) = 0$ $w h e n x^{4} - 1 = 0$ $x^{4} = 1 = c o s 2 k π + i s i n 2 k π$ $x = c o s (\frac{2 k π}{4}) + i s i n (\frac{2 k π}{4}) = c o s (\frac{k π}{2}) + i s i n (\frac{k π}{2})$ $p u t k = 0, 1, 2, 3$ $x^{5} + 1 = 0$ $x^{5} = - 1 = c o s π + i s i n π = c o s (2 k π + π) + i s i n (2 k π + π)$ $x = c o s (\frac{2 k + 1}{5}) π + i s i n (\frac{2 k + 1}{5}) π$ $p u t n = 0, 1, 2, 3, 4$
	Commented by Tawa1 last updated on 02/Jan/19

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

	$8) {(x + 1)}^{6} + {(x - 1)}^{6} = 0$ $(x^{6} + 6 c_{1} x^{5} + 6 c_{2} x^{4} + 6 c_{3} x^{3} + 6 c_{4} x^{2} + 6 c_{5} x + 6 c_{6}) +$ $(x^{6} - 6 c_{1} x^{5} + 6 c_{2} x^{4} - 6 c_{3} x^{3} + 6 c_{4} x^{2} - 6 c_{5} x + 6 c_{6}) = 0$ $a d d i n g$ $2 (x^{6} + \frac{6 \times 5}{2} x^{4} + \frac{6 \times 5}{2} x^{2} + 1) = 0$ $x^{6} + 15 x^{4} + 15 x^{2} + 1 = 0$ $x^{3} + 15 x + \frac{15}{x} + \frac{1}{x^{3}} = 0$ $(x^{3} + \frac{1}{x^{3}}) + 15 (x + \frac{1}{x}) = 0$ ${(x + \frac{1}{x})}^{3} - 3 (x + \frac{1}{x}) + 15 (x + \frac{1}{x}) = 0$ ${(x + \frac{1}{x})}^{3} + 12 (x + \frac{1}{x}) = 0$ $(x + \frac{1}{x}) [{(x + \frac{1}{x})}^{2} + 12] = 0$ $w h e n x + \frac{1}{x} = 0$ $x^{2} + 1 = 0$ $x^{2} = - 1 = c o s π + i s i n π = c o s (2 k π + π) + i s i n (2 k π + π)$ $x = c o s (\frac{2 k + 1}{2}) π + i s i n (\frac{2 k + 1}{2}) π$ $p u t k = 0 a n d 1$ $x_{1} = c o s \frac{π}{2} + i s i n \frac{π}{2} = i$ $x_{2} = c o s (\frac{3 π}{2}) + i s i n (\frac{3 π}{2}) = - i$ ${(x + \frac{1}{x})}^{2} + 12 = 0$ $x^{2} + 2 + \frac{1}{x^{2}} + 12 = 0$ $x^{4} + 14 x^{2} + 1 = 0$ $x^{2} = \frac{- 14 \pm \sqrt{196 - 4}}{2} = \frac{- 14 \pm \sqrt{192}}{2} = \frac{- 14 \pm 8 \sqrt{3}}{2}$ $x^{2} = - 7 + 4 \sqrt{3} = - 1 (7 - 4 \sqrt{3})$ $x^{2} = i^{2} {(2 - \sqrt{3})}^{2} x = \pm i (2 - \sqrt{3})$ $w h e n x^{2} = - 7 - 4 \sqrt{3} = - 1 (7 + 4 \sqrt{3})$ $x^{2} = i^{2} {(2 + \sqrt{3})}^{2}$ $x = \pm i (2 + \sqrt{3})$ $s o r o o t s a r e x \to [\pm i, \pm i (2 - \sqrt{3}), \pm i (2 + \sqrt{3})]$ $a l t e r n a t i v e w a y t o s o l v e . . . . i a m t r y i n g . . .$
	Commented by Tawa1 last updated on 02/Jan/19

	$God bless you sir . Thanks for your time sir . i appreciate$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19

	$9) {(x + 1)}^{7} = {(x - 1)}^{7}$ $\frac{{(x + 1)}^{7}}{{(x - 1)}^{7}} = 1$ ${(\frac{x + 1}{x - 1})}^{7} = (1)$ ${(\frac{x + 1}{x - 1})}^{7} = c o s (2 k π) + i s i n (2 k π)$ $\frac{x + 1}{x - 1} = c o s (\frac{2 k π}{7}) + i s i n (\frac{2 k π}{7}) = r$ $x + 1 = r x - r$ $x - r x = - r - 1$ $r x - x = r + 1$ $x = \frac{r + 1}{r - 1} = \frac{c o s (\frac{2 k π}{7}) + i s i n (\frac{2 k π}{7}) + 1}{c o s (\frac{2 k π}{7}) + i s i n (\frac{2 k π}{7}) - 1}$ $x = \frac{2 {c o s}^{2} (\frac{k π}{7}) + i 2 s i n (\frac{k π}{7}) c o s (\frac{k π}{7})}{- 2 {s i n}^{2} (\frac{k π}{7}) + i 2 s i n (\frac{k π}{7}) c o s (\frac{k π}{7})}$ $x = c o t (\frac{k π}{7}) \times \frac{c o s \frac{k π}{7} + i s i n \frac{k π}{7}}{i c o s \frac{k π}{7} - s i n \frac{k π}{7}}$ $x = i c o t (\frac{k π}{7}) \times \frac{c o s \frac{k π}{7} + i s i n \frac{i k π}{7}}{- c o s \frac{k π}{7} - i s i n \frac{k π}{7}}$ $x = - i c o t (\frac{k π}{7}) \times \frac{c o s \frac{k π}{7} + i s i n \frac{k π}{7}}{c o s \frac{k π}{7} + i s i n (\frac{k π}{7})}$ $x = - i c o t (\frac{k π}{7})$
	Commented by Tawa1 last updated on 02/Jan/19

	$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jan/19

	$8) a l t e r n a t i v e . .$ ${(x + 1)}^{6} + {(x - 1)}^{6} = 0$ ${(\frac{x + 1}{x - 1})}^{6} + 1 = 0$ ${(\frac{x + 1}{x - 1})}^{6} = - 1 = c o s (2 k π + π) + i s i n (2 k π + π)$ $(\frac{x + 1}{x - 1}) = c o s (\frac{2 k π + π}{6}) + i s i n (\frac{2 k π + π}{6})$ $\frac{x + 1}{x - 1} = r$ $x + 1 = r x - r$ $x - r x = - r - 1$ $r x - x = r + 1$ $x = \frac{r + 1}{r - 1} = \frac{1 + \frac{1}{r}}{1 - \frac{1}{r}} = \frac{1 + c o s (\frac{2 k π + π}{6}) - i s i n (\frac{2 k π + π}{6})}{1 - c o s (\frac{2 k π + π}{6}) + i s i n (\frac{2 k π + π}{6})}$ $x = \frac{2 {c o s}^{2} (\frac{2 k π + π}{12}) - 2 i s i n (\frac{2 k π + π}{12}) c o s (\frac{2 k π + π}{12})}{2 {s i n}^{2} (\frac{2 k π + π}{12}) + 2 i c o s (\frac{2 k π + π}{12}) s i n (\frac{2 k π + π}{12})}$ $x = c o t (\frac{2 k π + π}{12}) \times \frac{c o s (\frac{2 k π + π}{12}) - i s i n (\frac{2 k π + π}{12})}{s i n (\frac{2 k π + π}{12}) + i c o s (\frac{2 k π + π}{12})}$ $x = i c o t (\frac{2 k π + π}{12}) \times \frac{c o s (\frac{2 k π + π}{12}) - i s i n (\frac{2 k π + π}{12})}{- c o s (\frac{2 k π + π}{12}) + i s i n (\frac{2 k π + π}{12})}$ $x = - i c o t (\frac{2 k π + π}{12})$
	Commented by Tawa1 last updated on 03/Jan/19

	$God bless you sir$
	Commented by Tawa1 last updated on 03/Jan/19

	$Sir, what of the deduction sir . Thanks for your time sir .$
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