Given, x² - x + 1 = 0

Now, by Shridharacharya formula, we het

      x = {1 ± √(1 - 4*1*1) }/2

=> x = {1 ± √(1 - 4) }/2

=> x = {1 ± √(-3)}/2

=> x = {1 ± √(3 * -1)}/2

=> x = {1 ± √3 * √-1}/2

=> x = {1 ± i√3}/2                            {since i = √-1}

=> x = {1 + i√3}/2, {--1 - i√3}/2

=> x = -{-1 - i√3}/2, -{-1 + i√3}/2

=> x = w, w²                                         {since w =  {-1 + i√3}/2 and w² = {-1 - i√3}/2 }

Hence, α = -w, β = w²

Again we know that w³ = 1 and 1 + w + w² = 0

Now, α²⁰⁰⁹ + β²⁰⁰⁹ = α²⁰⁰⁷ * α² + β²⁰⁰⁷ * β²

                                = (-w)²⁰⁰⁷ * (-w)² + (-w² )²⁰⁰⁷ * (-w² )²            {since 2007 is multiple of 3}

                                = -(w)²⁰⁰⁷ * (w)² - (w² )²⁰⁰⁷ * (w⁴ )

                                = -1 * w² - 1 * w³ * w

                                = -1 * w² - 1 * 1 * w

                                = -w² - w

                                = 1                                  {since 1 + w + w² = 0}

So, α²⁰⁰⁹ + β²⁰⁰⁹ = 1