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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}

See also

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