if x is a cube root of a unity prove that (1−x)^6 =−27


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Question Number 111624 by mathdave last updated on 04/Sep/20

$i f x i s a c u b e r o o t o f a u n i t y$ $p r o v e t h a t$ ${(1 - x)}^{6} = - 27$
	Answered by Her_Majesty last updated on 04/Sep/20

	${(1 - ω)}^{6} = ω^{6} - 6 ω^{5} + 15 ω^{4} - 20 ω^{3} + 15 ω^{2} - 6 ω + 1 =$ $= 1 - 6 ω^{2} + 15 ω - 20 + 15 ω^{2} - 6 ω + 1 =$ $= 9 (ω + ω^{2}) - 18 = - 27$
	Answered by $@y@m last updated on 04/Sep/20
	$(1−x)^6 ={(1−x)^3 }^2 ={1−x^3 −3x(1−x)}^2 ={1−1−3x(1−x)}^2 =9x^2 (1−2x+x^2 ) =9(x^2 −2x^3 +x^4 ) =9(x^2 −2+x) =9(−2−1) =−27$
	${(1 - x)}^{6} = {{(1 - x)}^{3}}^{2}$ $= {1 - x^{3} - 3 x (1 - x)}^{2}$ $= {1 - 1 - 3 x (1 - x)}^{2}$ $= 9 x^{2} (1 - 2 x + x^{2})$ $= 9 (x^{2} - 2 x^{3} + x^{4})$ $= 9 (x^{2} - 2 + x)$ $= 9 (- 2 - 1)$ $= - 27$
	Commented by Rasheed.Sindhi last updated on 04/Sep/20

	$C o o l!$
	Commented by $@y@m last updated on 04/Sep/20

	$T h a n k s$
	Answered by mathdave last updated on 04/Sep/20

	$s o l u t i o n$ $l e t y^{3} = 1$ $y = {(1)}^{\frac{1}{3}} = {(\cos (0) + j \sin (0))}^{\frac{1}{3}} t o f i n d t h e r o o t s l e t$ $y = {(\cos 2 n π + j \sin 2 n π)}^{\frac{1}{3}}$ $y = (\cos \frac{2 n π}{3} + j \sin \frac{2 n π}{3}) p u t t i n g n = 0, 1, 2 w h i c h a r e t h e r o o t s o f t h e u n i t y$ $a t y_{n} = y_{0} = 1 . . . . . . . . . . (1)$ $y_{1} = {[\cos \frac{2 π}{3} + j \sin \frac{2 π}{3}]}^{1} = x . . . . . . . . . . (2), a t$ $y_{2} = {[\cos \frac{4 π}{3} + j \sin \frac{4 π}{3}]}^{2} = x^{2} . . . . . . . . (3)$ $a d d i n g (1), (2), a n d (3)$ $1 + x + x^{2} = 1 + \cos \frac{2 π}{3} + j \sin \frac{2 π}{3} + \cos \frac{4 π}{3} + j \sin \frac{4 π}{3}$ $1 + x + x^{2} = 1 + \cos (π - \frac{π}{3}) + j \sin (π - \frac{π}{3}) + \cos (π + \frac{π}{3}) + j \sin (π + \frac{π}{3})$ $1 + x + x^{2} = 1 - \cos \frac{π}{3} + j \sin \frac{π}{3} - \cos \frac{. π}{3} - j \sin \frac{π}{3}$ $1 + x + x^{2} = 1 - 2 \cos \frac{π}{3} = 1 - 2 (\frac{1}{2}) = 0$ $1 + x + x^{2} = 0$ $1 + x^{2} = - x . . . . . . . . . (1)$ $b u t {(1 - x)}^{6} = {[{(1 - x)}^{2}]}^{3}$ ${(1 - x)}^{6} = {(1 - 2 x + x^{2})}^{3} = {[(1 + x^{2}) - 2 x]}^{3}$ ${(1 - x)}^{6} = {(- x - 2 x)}^{3} = {(- 3 x)}^{3}$ ${(1 - x)}^{6} = - 27 x^{3}$ $b u t x^{3} = 1$ $∵ {(1 - x)}^{6} = - 27 Q . E . D$
	Commented by Tawa11 last updated on 06/Sep/21

	$great sir$
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