Question

(1+2omega+3omega^2)(3+2omega+omega^2)

Answer 1

I have conclude it till 3+15omega+10omega^2+8omega^3

Answer 2

Or 11+15omega+10omega^2

Answer 3

write 1+2O+3O^2 = 1+O+O^2 + O+2O^2
that would simplify it

Answer 4

as 1+O+O^2 =0

Answer 5

1 + w + w^2 can be equated to zero
hence
ur left with
(w + 2w^2) (2+w)
=2w + w^2 +4w^2 +2w^3
now simplify similarly

Answer 6

yeah, just like that ^

Answer 7

w^3 = 1

Answer 8

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

Answer 9

u can simplify it more
1+2O+3O^2 = 1+O+O^2 + O+2O^2 = O+2O^2 = O^2 + (-O)

Answer 10

And we know that 1+w+w^2=0

so,

(w+2w^2) (3+2w+w^2 ) Right @hartnn

Answer 11

yeah
do it same for other bracket

Answer 12

More over we know that w+w^2 =-1

so,
(w+2w^2)(3+w-1)
(w+2w^2)(2+w)

Answer 13

@hartnn

Answer 14

yeah, go ahead

Answer 15

2w+w^2+4w^2+2w^3

2w+5w^2+2w^3

2w+2w^3+5w^2

2(w+1)+5w^2

2(-w^2)+5w^2

Therfore the final expression we get is,

3w^2-----Answer :) @hartnn

Answer 16

yeah

Answer 17

Thanks Alot :)