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Question: If $\alpha$, $\beta \in C$ are the distinct roots, of the equation $x^2 - x + 1 = 0$, then $\alpha^{...

If α\alpha, β∈C\beta \in C are the distinct roots, of the equation x2−x+1=0x^2 - x + 1 = 0, then α101+β107\alpha^{101} + \beta^{107} is equal to

A

2

B

-1

C

0

D

1

Answer

1

Explanation

Solution

The given equation is x2−x+1=0x^2 - x + 1 = 0. Multiplying by (x+1)(x+1), we get (x+1)(x2−x+1)=0(x+1)(x^2 - x + 1) = 0, which simplifies to x3+1=0x^3 + 1 = 0. Thus, x3=−1x^3 = -1 for the roots α\alpha and β\beta. This implies α3=−1\alpha^3 = -1 and β3=−1\beta^3 = -1. Squaring these, we get α6=1\alpha^6 = 1 and β6=1\beta^6 = 1.

To find α101\alpha^{101}: 101=16×6+5101 = 16 \times 6 + 5. So, α101=α16×6+5=(α6)16⋅α5=116⋅α5=α5\alpha^{101} = \alpha^{16 \times 6 + 5} = (\alpha^6)^{16} \cdot \alpha^5 = 1^{16} \cdot \alpha^5 = \alpha^5. Since α3=−1\alpha^3 = -1, α5=α3⋅α2=−1⋅α2=−α2\alpha^5 = \alpha^3 \cdot \alpha^2 = -1 \cdot \alpha^2 = -\alpha^2. From the original equation, α2−α+1=0\alpha^2 - \alpha + 1 = 0, so α2=α−1\alpha^2 = \alpha - 1. Therefore, α101=−(α−1)=1−α\alpha^{101} = -(\alpha - 1) = 1 - \alpha.

To find β107\beta^{107}: 107=17×6+5107 = 17 \times 6 + 5. So, β107=β17×6+5=(β6)17⋅β5=117⋅β5=β5\beta^{107} = \beta^{17 \times 6 + 5} = (\beta^6)^{17} \cdot \beta^5 = 1^{17} \cdot \beta^5 = \beta^5. Since β3=−1\beta^3 = -1, β5=β3⋅β2=−1⋅β2=−β2\beta^5 = \beta^3 \cdot \beta^2 = -1 \cdot \beta^2 = -\beta^2. From the original equation, β2−β+1=0\beta^2 - \beta + 1 = 0, so β2=β−1\beta^2 = \beta - 1. Therefore, β107=−(β−1)=1−β\beta^{107} = -(\beta - 1) = 1 - \beta.

Now, summing the results: α101+β107=(1−α)+(1−β)=2−(α+β)\alpha^{101} + \beta^{107} = (1 - \alpha) + (1 - \beta) = 2 - (\alpha + \beta). From Vieta's formulas for x2−x+1=0x^2 - x + 1 = 0, the sum of the roots α+β=−(−1)/1=1\alpha + \beta = -(-1)/1 = 1. Thus, α101+β107=2−1=1\alpha^{101} + \beta^{107} = 2 - 1 = 1.