# Specimen Cyprus Seniors Provincial/2nd grade/Problem 4

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## Problem

If $\rho_{1}, \rho_{2}$ are the roots of equation $x^2-x+1=0$ then:

a) Prove that $\rho_{1}^3=\rho_{2}^3 = -1$ and

b) Calculate the value of: $\rho_{1}^{2006} + \rho_{2}^{2006}$.

## Solution

Since $\rho_1, \rho_2 \neq -1$, we can multiply both sides by $x+1$, and $\rho_{1}, \rho_{2}$ will still satisfy the equation:

$(x^2-x+1)(x+1)=(0)(x+1)$

$x^3 + 1 = 0$

$x^3 = -1$

Thus, $\rho_{1}^3=\rho_{2}^3 = -1$ as desired $\boxed{\mathbb{Q.E.D.}}$.

The third roots of $-1 = cis(180^\circ)$ are $cis(60^\circ)$, $cis(-60^\circ)$, and $-1$.

Since $\rho_1, \rho_2 \neq -1$, we have $\rho_{1} = cis(60^\circ)$ and $\rho_{2} = cis(-60^\circ)$. (Switching the two values will not affect the result).

Note that $\rho_{1}^6=\rho_{2}^6 = 1$, and that $2006 = 6(334)+2$

So, $\rho_{1}^{2006} + \rho_{2}^{2006} = \rho_{1}^{2} + \rho_{2}^{2} = cis(120^\circ) + cis(-120^\circ) =$ $\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \boxed{-1}$