Key Concepts and Formulas

• Roots of $x^2 - x + 1 = 0$ satisfy $x^3 = -1$.
• Vieta's formulas for a quadratic equation $ax^2 + bx + c = 0$: Sum of roots = $-\frac{b}{a}$, Product of roots = $\frac{c}{a}$.
• Algebraic identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

Step-by-Step Solution

Step 1: Simplify the High Powers of the Roots

Our goal is to simplify $\alpha^{101} + \beta^{107}$ using the fact that $\alpha$ and $\beta$ are roots of $x^2 - x + 1 = 0$, and thus $\alpha^3 = -1$ and $\beta^3 = -1$.

Why this step? Direct computation of high powers is difficult. Using the property $x^3=-1$ simplifies the exponents through modular arithmetic.

We divide the exponents by 3: $101 = 3 \times 33 + 2$ $107 = 3 \times 35 + 2$

Then, we rewrite the powers: $\alpha^{101} = \alpha^{3 \times 33 + 2} = (\alpha^3)^{33} \cdot \alpha^2 = (-1)^{33} \cdot \alpha^2 = - \alpha^2$ $\beta^{107} = \beta^{3 \times 35 + 2} = (\beta^3)^{35} \cdot \beta^2 = (-1)^{35} \cdot \beta^2 = - \beta^2$

Therefore, $\alpha^{101} + \beta^{107} = - \alpha^2 - \beta^2 = -(\alpha^2 + \beta^2)$

Step 2: Evaluate the Sum of Squares Using Vieta's Formulas

We need to find $\alpha^2 + \beta^2$. We'll use Vieta's formulas to express it in terms of the sum and product of the roots.

Why this step? Vieta's formulas provide a direct relationship between roots and coefficients without explicitly calculating the roots.

For the quadratic $x^2 - x + 1 = 0$, $a = 1$, $b = -1$, and $c = 1$. Vieta's formulas give us: $\alpha + \beta = -\frac{b}{a} = -\frac{-1}{1} = 1$ $\alpha \beta = \frac{c}{a} = \frac{1}{1} = 1$

We can rewrite $\alpha^2 + \beta^2$ using the identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

Why this step? This is a standard algebraic identity that allows us to use the results from Vieta's formulas.

Substituting the values we found: $\alpha^2 + \beta^2 = (1)^2 - 2(1) = 1 - 2 = -1$

Step 3: Final Calculation

Now we substitute the value of $\alpha^2 + \beta^2$ back into our expression:

$\alpha^{101} + \beta^{107} = -(\alpha^2 + \beta^2) = -(-1) = 1$

Common Mistakes & Tips

• Remember that roots of $x^2 - x + 1 = 0$ satisfy $x^3 = -1$, and roots of $x^2 + x + 1 = 0$ satisfy $x^3 = 1$.
• Be careful with signs when dealing with powers of $-1$.
• Vieta's formulas are essential for quickly finding the sum and product of roots.

Summary

We simplified the expression $\alpha^{101} + \beta^{107}$ by using the fact that $\alpha$ and $\beta$ are roots of $x^2 - x + 1 = 0$, implying $\alpha^3 = \beta^3 = -1$. This allowed us to reduce the exponents. We then used Vieta's formulas to find the sum and product of the roots, and the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ to calculate $\alpha^2 + \beta^2$. Finally, we substituted this value back into our simplified expression to find the result.

The final answer is \boxed{1}, which corresponds to option (D).