A student believed that T is a root of x^4 - x^3 + x^2 - x + 1 = 0. Find the value of T^20 + T^10 + T^5 +1.

The values of the roots for the given equation are `root[5](-1),-(-1)^(2/5),(-1)^(3/5),-(-1)^(4/5)` . Whichever value is selected as T, we have `T^(20)+T^(10)+T^(5)+1=2` .

Expert Answers

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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.

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