1.

If `x^2+x+1=0` then the value of `(x+1/x)^2+(x^2+1/(x^2))^2+...+(x^27+1/(x^27))^2` isA. 27B. 72C. 45D. 54

Answer» Correct Answer - D
`x^(2)+x+1=0`
`rArr x = omegaor omega^(2)`
Let `x = omega`. Then,
`x+(1)/(x)=omega+(1)/(omega)=omega+omega^(2)=-1`
`x^(2)+(1)/(x^(2))=omega^(2)+(1)/(omega^(3))=omega^(2)+omega=-1`
`x^(3)+(1)/(x^(3))=omega^(3)+(1)/(omega^(3))=2`
`x^(4)+(1)/(x^(4))=omega^(4)+(1)/(omega^(4))=omega+omega^(2)=-1,` etc.
`therefore (x+(1)/(x))^(2)+(x^(2)+(1)/(x^(2)))^(2)+(x^(3)+(1)/(x^(3)))^(2)+* * *+(x^(27)+(1)/(x^(27)))^(2)`
`=[(x+(1)/(x))^(2)+(x^(2)+(1)/(x^(2)))^(2)+(x^(4)+(1)/(x^(4)))^(2)+* * *+(x^(26)+(1)/(x^(26)))^(2)]`
`+[(x^(3)+(1)/(x^(3)))^(2)+(x^(6)+(1)/(x^(6)))^(2)+(x^(9)+(1)/(x^(9)))^(2)+* * *+(x^(27)+(1)/(x^(27)))^(2)]`
`=18+9(2)^(2)=54`


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