If w is complex cube root of unity, then w-16
=
a)w
b)1
c)w2
d)w-1
If w is complex cube root of unity, then w-27
=
a)w
b) 1
c)w2
d) w
Cube roots of 8 are:
a) 2,2w,2w2
b) -2,-2w,-2w2
c) 2,-2w,-2w2
d) 2,-2w,2w2
Cube roots of -27 are:
a) -3,-3w,-3w2
b) -2,-2w,-2w2
c) 3,-2w,-3w2
d) 2,-2w,2w2
Cube roots of 64 are:
a) -4,-4w,-4w2
b) -2,-4w,-2w2
c) 4,4w,4w2
d) 2,-2w,2w2
(1-w-w2)5=…………
a) 16
b) 64
c) 6
d)32
Solution:(1-w-w2)5
(1-(w+w2))5 As, 1+w+w2=0
Or w+w2=-1 = (1-(-1))5
(1+1)5 = (2)5 =32
