math
posted by Praphul .
Show that 1/2+ 1/2(i3 root) is a cube root of 1

Did you mean
show that 1/2 + (1/2)√3 i is a cube root of 1 ??
If so, then show that (1/2 + (1/2)√3 i )^3 = 1
LS = [ (1/2) ( 1  √3i) ]^3
= (1/2)^3 [ 1 + 3(√3i) + 3(√3i)^2 + (√3i)^3 ]
= (18) [ 1  3√3i + 9i^2  3√3 i^3 ]
= (1/8)(1  3√3  9 + 3√3i ]
=(1/8)[8}
= 1
= RS
Yes, 1/2 + (1/2)√3 i is a cube root of 1
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