If `omega!=1` is a cube root of unity and `x+y+z!=0,` then prove that `|[x/(1+omega), y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[(z)/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|=0` if `x=y=z`

Options

A. `x^(2)+y^(2)+z^(2)=0`
B. `x+yomega+zomega^(2)=0` or `x=y=z`
C. `x ne y ne z ne 0`
D. x=2y=3z

Answer

B

Explanation

`(b)` As `1+omega+omega^(2)=0`
`D=|{:(-(x)/(omega^(2)),-y,-(z)/(omega)),(-y,-(z)/(omega),-(x)/(omega^(2))),(-(z)/(omega),-(x)/(omega^(2)),-y):}|=x^(3_+y^(3)+z^(3)-3xyz`
`=(1)/(2)(x+y+z){(x-y)^(2)+(y-z)^(2)+(z-x)^(2)}`
`=(x+y+z)(x+yomega+zomega^(2))(x+yomega^(2)+zomega)`
The determinant varnishes if `x=y=z` or `x+yomega+zomega^(2)=0`