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` is

Options

A. 27
B. 72
C. 45
D. 54

Answer

D

Explanation

We have,
`x^(2)+x+1=0 rArr x = omega` and `omega^(2)`
Let `x = omega`, Then,
`x+1/x=omega+1/omega^(2)=omega+omega^(2)=-1`
`x^(2)+1/x^(2)=omega^(2)=omega^(2)+omega=1`
`x^(3)+1/x^(3)=omega^(3)+1/omega^(3)=2`
In fact, `x^(n)+1/x^(n) = {{:(-1,"if n" ne 3k),(2,"if n"=3k):}`
`therefore (x+1/x)^(2)+(x^(2)+1/x^(2))^(2) + (x^(3)+1/x^(3))^(2)+............+(x^(27)+1/x^(27))^(2)`
`=(a+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`