Question

Prove that w^49+w^101+w^150=0,where w is a complex cube root of unity

Answer 1

now this is simple

Answer 2

w^3 =1
did u know this ?

Answer 3

REALLY!

Answer 4

-1?

Answer 5

yes i know w^3 =1 ,w+w^2+1=0 and etc

Answer 6

so w^49 = ?

Answer 7

w^49 = w^48 . w = (w^3)^16 . w = ??

Answer 8

w

Answer 9

similarly try to find w^101

Answer 10

ok wait

Answer 11

w^101=> w^100.w => (w^3)^33.w.w=>w^2

Answer 12

good :)
w^150 = ?

Answer 13

w^150 => (w^3)^50 => 1

Answer 14

so?

Answer 15

w^49+w^101+w^150=?

Answer 16

w^2+w+1=0

Answer 17

thanks

Answer 18

0

Answer 19

welcome :)