If alpha and beta are roots of the equation x | vedclass

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(B) The given equation is $x^2 - x + 1 = 0$.Multiplying by $(x + 1)$,we get $(x + 1)(x^2 - x + 1) = 0$,which implies $x^3 + 1 = 0$,so $x^3 = -1$.Since

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$1$

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(B) The given equation is $x^2 - x + 1 = 0$.Multiplying by $(x + 1)$,we get $(x + 1)(x^2 - x + 1) = 0$,which implies $x^3 + 1 = 0$,so $x^3 = -1$.Since $\alpha$ and $\beta$ are roots of the equation,they satisfy $x^3 = -1$,so $\alpha^3 = -1$ and $\beta^3 = -1$.We need to find $\alpha^{2009} + \beta^{2009}$.$\alpha^{2009} = (\alpha^3)^{669} \cdot \alpha^2 = (-1)^{669} \cdot \alpha^2 = -\alpha^2$.Similarly,$\beta^{2009} = -\beta^2$.Thus,$\alpha^{2009} + \beta^{2009} = -(\alpha^2 + \beta^2)$.From the equation $x^2 - x + 1 = 0$,we have $\alpha + \beta = 1$ and $\alpha\beta = 1$.We know $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1)^2 - 2(1) = 1 - 2 = -1$.Therefore,$\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

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

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