If $\alpha \quad and\quad \beta$ are the roots of the equation ${ x }^{ 2 }-x+1=0,\quad then\quad { \alpha }^{ 2009 }+{ \beta }^{ 2009 }$

For the quadratic equation x^2 - x + 1 = 0, the sum of the roots is 1 and the product is 1. The roots satisfy the relation alpha^2 = alpha - 1 and beta^2 = beta - 1. Multiplying by x and substituting repeatedly shows the roots follow a cycle where alpha^3 = -1 and beta^3 = -1. Since 2009 is equal to 3 times 669 plus 2, we have alpha^2009 = (alpha^3)^669 multiplied by alpha^2, which simplifies to (-1)^669 multiplied by (alpha - 1), resulting in -alpha + 1. Applying the same logic gives beta^2009 = -beta + 1, so adding them together yields 2 - (alpha + beta) = 2 - 1 = 1.

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