If $\alpha$ and $\beta$ are the distinct roots of the equation $x^2 - x + 1 = 0$, then the value of $\alpha^{200} + \beta^{206} + 2$ is equal to

Given equation: $x^2 - x + 1 = 0$

Multiplying both sides by $(x + 1)$, we get:

$(x + 1)(x^2 - x + 1) = 0$

$x^3 + 1 = 0 \Rightarrow x^3 = -1$

Since $\alpha$ and $\beta$ are roots of $x^2 - x + 1 = 0$, they satisfy the equation $x^3 = -1$.

Therefore, $\alpha^3 = -1$ and $\beta^3 = -1$.

Now, evaluating the given expression $\alpha^{200} + \beta^{206} + 2$ :

$\alpha^{200} = (\alpha^3)^{66} \cdot \alpha^2 = (-1)^{66} \cdot \alpha^2 = \alpha^2$

$\beta^{206} = (\beta^3)^{68} \cdot \beta^2 = (-1)^{68} \cdot \beta^2 = \beta^2$

Substituting these values, the expression becomes $\alpha^2 + \beta^2 + 2$.

From the given quadratic equation, the sum of roots $\alpha + \beta = 1$ and the product of roots $\alpha \beta = 1$.

Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ :

$\alpha^2 + \beta^2 = (1)^2 - 2(1) = -1$

Therefore, $\alpha^{200} + \beta^{206} + 2 = -1 + 2 = 1$.