Question

If \(α\) and \(β\) are the distinct roots of the equation \(x^2-x+1 = 0\), then the value of \(α^{200}+β^{206}+2\) is equal to

• A. \(1\)
• B. \(-1\)
• C. \(0\)
• D. \(2\)

Hint:

The roots satisfy \(\alpha^6=\beta^6=1\), so reduce each power modulo 6.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The roots of \(x^2-x+1=0\) are \(\alpha,\beta = \dfrac{1\pm i\sqrt3}{2} = e^{\pm i\pi/3}\). Their sixth power is \(e^{\pm 2\pi i}=1\).

Step 2: Key Formula or Approach
From the equation: \(\alpha+\beta = 1\) and \(\alpha\beta = 1\). Also \(\alpha^6=\beta^6=1\), so only the remainder of the power on division by 6 matters.

Step 3: Detailed Explanation
\(200 = 6\times33+2\), so \(\alpha^{200}=\alpha^2\).
\(206 = 6\times34+2\), so \(\beta^{206}=\beta^2\).
\[ \alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = 1-2 = -1 \]
\[ \alpha^{200}+\beta^{206}+2 = -1+2 = 1 \]

Final Answer:
The value is 1, option (A). \[ \boxed{1\ \text{(A)}} \]