If `x^2-x+1=0` then the value of `sum_[n=1]^[5][x^n+1/x^n]^2` is:

Options

A. 8
B. 10
C. 12
D. none of these

Answer

A

Explanation

We have,
`x^(2)-x+1=0 rArr x=-omega-omega^(2)`
For `x=-omega`, we have
`sum_(n=1)^(5)(x^(n)+1/x^(n))^(2)`
`=sum_(n=1)^(5){(-omega)^(n)+1/(-omega)^(n)}^(2)`
`=sum_(n=1,n in 3)^(5)(omega^(n)+omega^(2n))^(2)+(omega^(3)+omega^(6))^(2)`
`=(sum_(n=1), n in 3)^(5) (-1)^(2)+(2)^(2)`
`4+4=8`
Similarly, for `x=-omega^(2)`, we obtain that the value of the given expression is 8.