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

The quadratic $$x^{2}-x+1=0$$ has roots $$\alpha,\beta$$.

By Viète’s relations, $$\alpha+\beta = 1$$ and $$\alpha\beta = 1$$.

Since each root satisfies the equation, we can write for every integer $$n \ge 2$$

$$\alpha^{n}= \alpha^{n-1}-\alpha^{n-2}, \quad \beta^{n}= \beta^{n-1}-\beta^{n-2}$$

Add the two relations and define $$S_{n}= \alpha^{n}+\beta^{n}$$. This gives the linear recurrence

$$S_{n}=S_{\,n-1}-S_{\,n-2} \qquad -(1)$$

Initial terms:

$$S_{0}= \alpha^{0}+\beta^{0}=2,$$ $$S_{1}= \alpha+\beta = 1.$$

Use (1) to generate the next four terms:

$$S_{2}=S_{1}-S_{0}=1-2=-1,$$ $$S_{3}=S_{2}-S_{1}=-1-1=-2,$$ $$S_{4}=S_{3}-S_{2}=-2-(-1)=-1,$$ $$S_{5}=S_{4}-S_{3}=-1-(-2)=1.$$

Repeating the step once more gives $$S_{6}=S_{5}-S_{4}=1-(-1)=2,$$ so the sequence $$S_{n}$$ is periodic with period $$6$$:

$$\;n:\;0\;1\;2\;3\;4\;5\;6\;7\ldots$$ $$S_{n}: 2,1,-1,-2,-1,1,2,1,\ldots$$

Now reduce the exponent modulo the period: $$2009 \div 6 = 334$$ remainder $$5$$, so $$2009 \equiv 5 \pmod{6}$$.

Therefore $$S_{2009}=S_{5}=1$$, i.e.

$$\alpha^{2009}+\beta^{2009}=1.$$

Option B which is: $$1$$