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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 roots α,β\alpha, \beta of the equation x2−x+1=0x^2 - x + 1 = 0 satisfy α3=−1\alpha^3 = -1 and β3=−1\beta^3 = -1. This is because multiplying the equation by (x+1)(x+1) yields (x+1)(x2−x+1)=x3+1=0(x+1)(x^2 - x + 1) = x^3 + 1 = 0.

We need to evaluate α101+β107\alpha^{101} + \beta^{107}. For α101\alpha^{101}: 101=3×33+2101 = 3 \times 33 + 2. So, α101=(α3)33⋅α2=(−1)33⋅α2=−1⋅α2=−α2\alpha^{101} = (\alpha^3)^{33} \cdot \alpha^2 = (-1)^{33} \cdot \alpha^2 = -1 \cdot \alpha^2 = -\alpha^2.

For β107\beta^{107}: 107=3×35+2107 = 3 \times 35 + 2. So, β107=(β3)35⋅β2=(−1)35⋅β2=−1⋅β2=−β2\beta^{107} = (\beta^3)^{35} \cdot \beta^2 = (-1)^{35} \cdot \beta^2 = -1 \cdot \beta^2 = -\beta^2.

Therefore, α101+β107=−α2−β2=−(α2+β2)\alpha^{101} + \beta^{107} = -\alpha^2 - \beta^2 = -(\alpha^2 + \beta^2).

From Vieta's formulas for the equation x2−x+1=0x^2 - x + 1 = 0: Sum of roots: α+β=1\alpha + \beta = 1 Product of roots: αβ=1\alpha \beta = 1

We know that α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta. Substituting the values: α2+β2=(1)2−2(1)=1−2=−1\alpha^2 + \beta^2 = (1)^2 - 2(1) = 1 - 2 = -1.

Finally, substituting this back into the expression for α101+β107\alpha^{101} + \beta^{107}: α101+β107=−(α2+β2)=−(−1)=1\alpha^{101} + \beta^{107} = -(\alpha^2 + \beta^2) = -(-1) = 1.