Question

Prove that (1-w-w^2)^6=64

Answer 1

U mean solve?

Answer 2

yup

Answer 3

i thin u can find what is w+w^2=??

Answer 4

(+-2)^6=64

Answer 5

Or u can use that

Answer 6

how???

Answer 7

1+w+w^2=0
what is w+w^2 ??

Answer 8

where is w+w^2 applying

Answer 9

(1-w-w^2)^6=64
(1-w-w^2)^6=(-+2)^6
1-w-w^2=-+2

Answer 10

Now, can u do it from here?

Answer 11

http://www.wolframalpha.com/input/?i=%281-w-w%5E2%29%5E6%3D64&x=0&y=0

Answer 12

i think its given that w is the cube root of unity, isn't it ?

Answer 13

(1-w-w^2)^6 should come 64

Answer 14

not solving we have to prove

Answer 15

1+w+w^2=0
so
w+w^2= -1
LEFT SIDE
= (1-(w+w^2))^6
=(1-(-1))^6
=2^6
=64
=RIGHT SIDE

Answer 16

yeah that was i talking about