We are given that T is a root of `x^4-x^3+x^2-x+1=0` , and we are asked to find the value of `T^(20)+T^(10)+T^(5)+1` .

To find the roots, we notice that the given expression is one of the factors of `x^5+1` , where the other factor is (x+1). So we multiply both sides of the given equation by (x+1) (introducing the extraneous solution x=-1).

`(x+1)(x^4-x^3+x^2-x+1=(x+1)0`

`x^5+1=0`

`x^5=-1` where `x ne -1` . Thus we find the complex roots (ignoring the real root at x=-1). The complex roots are `root[5](-1),-(-1)^(2/5), (-1)^(3/5)," and "-(-1)^(4/5)` , or alternatively `T=i^(2/5),-(i)^(4/5),i^(6/5),-(i)^(8/5)` , where `i` is the imaginary unit.

`(i^(2/5))^(20)+(i^(2/5))^(10)+(i^(2/5))^5+1=1+1-1+1=2`

`(-(i)^(4/5))^(20)+(-(i)^(4/5))^(10)+(-(i)^(4/5))^5+1=1+1-1+1=2`

The other two solutions also yield a sum of 2.

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now