if-x-is-a-cube-root-of-a-unity-prove-that-1-x-6-27-

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} - 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 \dots \dots \dots . (1)

y_{1} = {[\cos \frac{2 π}{3} + j \sin \frac{2 π}{3}]}^{1} = x \dots \dots \dots . (2), a t

y_{2} = {[\cos \frac{4 π}{3} + j \sin \frac{4 π}{3}]}^{2} = x^{2} \dots \dots . . (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 \dots \dots \dots (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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