Question

If α, β ∈ C are the distinct roots, of the equation x² − x + 1 = 0, then α¹⁰¹ + β¹⁰⁷ is equal to

• A. 2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (D)

1

Answer 2

The given equation is $x^2 - x + 1 = 0$. The roots of this equation are related to the complex cube roots of $-1$. Multiplying the equation by $(x+1)$, we get $(x+1)(x^2 - x + 1) = x^3 + 1 = 0$. Thus, the roots $\alpha$ and $\beta$ satisfy $\alpha^3 = -1$ and $\beta^3 = -1$. This also implies $\alpha^6 = 1$ and $\beta^6 = 1$.

We need to find $\alpha^{101} + \beta^{107}$. We can simplify the exponents using the property that powers repeat every 6. For $\alpha^{101}$: $101 \div 6$ gives a remainder of $5$ ($101 = 16 \times 6 + 5$). So, $\alpha^{101} = \alpha^5$. For $\beta^{107}$: $107 \div 6$ gives a remainder of $5$ ($107 = 17 \times 6 + 5$). So, $\beta^{107} = \beta^5$.

Therefore, $\alpha^{101} + \beta^{107} = \alpha^5 + \beta^5$.

Using $\alpha^3 = -1$, we have $\alpha^5 = \alpha^3 \cdot \alpha^2 = -1 \cdot \alpha^2 = -\alpha^2$. Similarly, $\beta^5 = \beta^3 \cdot \beta^2 = -1 \cdot \beta^2 = -\beta^2$.

So, $\alpha^5 + \beta^5 = -\alpha^2 - \beta^2 = -(\alpha^2 + \beta^2)$.

From the original equation $x^2 - x + 1 = 0$, by Vieta's formulas, the sum of the roots is $\alpha + \beta = -(-1)/1 = 1$, and the product of the roots is $\alpha\beta = 1/1 = 1$.

We can find $\alpha^2 + \beta^2$ using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1)^2 - 2(1) = 1 - 2 = -1$.

Substituting this back, we get $\alpha^5 + \beta^5 = -(-1) = 1$. Thus, $\alpha^{101} + \beta^{107} = 1$.