Question

Question #3bc5d

Answer 1

The answer is `16omega`.

Explanation:

Since, `root(3)(1)=omega`,
So, `omega^3=1`.-----(1).

`:.omega^3-1=0`
`:.(omega-1)(omega^2+omega+1)=0`

So, either first term is zero, or the second term is zero.

But, if `omega-1=0`, `omega=1`, which is not true.

So, `omega^2+omega+1=0`.
`:.omega^2+1=-omega`.----(2).

Now, `(3+omega+3omega^2)^4`
`={omega+3(omega^2+1)}^4`
`={omega+3(-omega)}^4` ----[From (2)].
`=(omega-3omega)^4`
`=(-2omega)^4`
`=16omega^4`
`=16omega^3xxomega`
`=16xx1xxomega` ----[From (1)].
`=16omega`. (Answer).

Hope it Helps :)