If w is a complex cube root of unity, then show that?

Question

(1 + 5w^2 + w^4) (1 + 5w + w^2) (5 + w + w^2) =64

Answer 1

Conside LHS:

Given, w is a complex cube root of unity (that is 1).

This implies w^3 = 1.



Also, w^4 = (w^3) (w) = (1) (w) = w



That is, w^4 = w

Also, the sum of the roots, 1 + w + w^2 = 0.



Separating w^2 & Substituting w^4 = w, we get,



(1 + 5w^2 + w^4) = (1+ w^2+4w^2+w)

= (1 + w + w^2 + 4w^2)

= (0 + 4w^2)

= 4w^2



(1 + 5w^2 + w^4) = 4w^2



Similarly,

(1 + 5w + w^2) = 4w &

(5 + w + w^2) = 4



Now, L.H.S. = (4w^2 ) (4w) (4)

= 64 w^3

= (64) (1)

= 64.

Thus Proved.......

Answer 2

if w is a complex cube root of unity:

w^3 = 1 (unity means 1 so w^3 must equal 1)

w^2 + w + 1 = 0 (sum of roots)



Using the above two:

LHS = (1 + 5w^2 + w^4) (1 + 5w + w^2) (5 + w + w^2)

= (1 + 5w^2 + w) (1 + 5w + w^2) (5 + w + w^2) , as w^3 = 1, w^4 = w

= (1 + w + w^2 + 4w^2) (1 + w + w^2 + 4w) (1 + w + w^2 + 4), separating out 1 + w + w^2 in each bracket

= (0 + 4w^2) (0 + 4w) (0 + 4), as 1 + w + w^2 = 0

= (4w^2) (4w) (4), getting rid of the zeroes

= 64w^3

= 64, as w^3 = 1

= RHS as required



hope this helped :)

Answer 3

Cube Roots Of Unity

Answer 4

(1 + 5w^2 + w^4) (1 + 5w + w^2) (5 + w + w^2)

= w^8 + 6w^7 + 16w^6 + 56w^5 + 61w^4 + 136w^3 + 36w^2 + 26w + 5

= w^2 + 6w + 16 + 56w^2 + 61w + 136 + 36w^2 + 26w + 5 ....... w^3 =1

= 93w^2 + 93w + 157

= 93(w^2 + w + 1) + 64

= 93(0) + 64

= 64

proved.

Answer 5

write 1=1 exp( i 2n pi) where n is an integer. Calculate the cube root cube root of 1= 1 exp( i 2npi/3) Put different n values. n=o root1=1 n=1 root2 = 1 exp(2i pi /3) =cos (120) + i sin(120) n=2 root3 = 1 exp(i 4 pi/3) = cos(240) + i sin (240) For n=3 onward the above roots repeat.Therefore these are distinct roots.

Answer 6

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Since w³ = 1, we have w³ - 1 = (w - 1)(w² + w + 1) = 0 ==> w² + w + 1 = 0, since w is complex (i.e., non-real). Hence, (1 + w - w²)³ - (1 - w + w²)³ = (1 + w - (-1 - w))³ - (1 - w + (-1 - w))³ = (2 + 2w)³ - (-2w)³ = 8 [(1 + w)³ + w³] = 8 [(1 + w) + w] [(1 + w)² - w(1 + w) + w²] = 8 (2w + 1) [(w² + 2w + 1) - (w + w²) + w²] = 8 (2w + 1) (w² + w + 1) = 16 (2w + 1) * 0 = 0. I hope this helps!

Answer 7

its not simply the cube root na..its complex cube root n i think it ll have 3 roots one of them is 1...i wanna ask what are the other ones?