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If omega!=1 is a cube root of unity and ...

If `omega!=1` is a cube root of unity and
`Delta=|(x+omega^(2),omega,1),(omega,omega^(2),1+x),(1,x+omega,omega^(2))|=0` then value of x is

A

`x=0`

B

`x=1`

C

`x=-1`

D

none of these

Text Solution

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The correct Answer is:
A
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