Question #3bc5d

1 Answer
Dec 16, 2016

Answer:

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 :)