Question

If W is a non real cube root of unity then \[(1+W)(1+w^2)(2+W^4)(1+W^8)(1+w^{10})(1+w^{32})\] =?

Answer 1

@mukushla @ghazi

Answer 2

\[w^4 = w
\]

Answer 3

w^8 = w^2 w^10 = w w^32 = w^2

Answer 4

(1+W)^3 + (1+w^2)^3

Answer 5

1+w + w^2 = 0

1+w = -w^2

Answer 6

(-w^2)^3 + (-w)^3

Answer 7

and...i am stuck from here

Answer 8

\[(1+w)(1+w^2)(1+w^4)(1+w^8)(1+w^{10})(1+w^{32})\]?

Answer 9

Find the value

Answer 10

\[w^2=-1-w\]\[w^4=w\]\[w^8=w^2=-1-w\]\[w^{10}=w\]\[w^{32}=w^2=-1-w\]

Answer 11

Yup..

Answer 12

i got this.....(-w^2)^3 + (-w)^3???

Answer 13

is that 2 there (2+w^4)?

Answer 14

1+w = -w^2

Answer 15

(1+W)^3 + (1+w^2)^3

Answer 16

\[(1+w)^2(1+w^2)^3(2+w)=(1+2w+w^2)(1-1-w)^3(2+w)\\=w\times(-w)^3\times(2+w)=-w(2+w)=-2w-w^2=-2w+1+w=1-w\]

Answer 17

but i think thats (1+W^4) and answer will be 1

Answer 18

yes the answer should be 1....

Answer 19

but hw did u get that

Answer 20

i got -2 as answer

Answer 21

so with my notations\[(1+w)(1+w^2)(1+w^4)(1+w^8)(1+w^{10})(1+w^{32})\\=(1+w)^3(1+w^2)^3\]right?

Answer 22

Yup...

Answer 23

\[(1+w)^3(1+w^2)^3=(1+w)(1+w)^2(1-1-w)^3\\=(1+w)(1+2w+w^2)(-w)^3=-w(1+w)=-w-w^2=1\]

Answer 24

sorry u r offline and i gave u the complete solution...see u yahoo