# Difference between revisions of "2007 UNCO Math Contest II Problems/Problem 4"

## Problem

If $x$ is a primitive cube root of one (this means that $x^3 =1$ but $x \ne 1$) compute the value of $$x^{2006}+\frac{1}{x^{2006}}+x^{2007}+\frac{1}{x^{2007}}.$$

## Solution

$\fbox{+1}$

Since $x^3=1$ ,$x^{2006}=x^2$ , $x^{2007}=1$, $x+\frac{1}{x}=-1$