Question

if Omega is a complex cube root of unity value bracket 1 - Omega plus omega square bracket close cube is

Answer 1

Answer

If ω is a complex cube root of unity, show that

⎛⎝⎜⎡⎣⎢1ωω2ωω21ω21ω⎤⎦⎥+⎡⎣⎢ωω2ωω21ω21ω1⎤⎦⎥⎞⎠⎟⎡⎣⎢1ωω2⎤⎦⎥=⎡⎣⎢000⎤⎦⎥

I tried to solve this and I reduced the L.H.S. to

⎡⎣⎢2+2ω+2ω22+2ω+2ω22+2ω+2ω2⎤⎦⎥

(since ω3=1)

but couldn't equate it to R.H.S.

Please provide your assistance.

Thank you

Answer 2

check the attachment

hope it helps