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: Use the polar form
$\alpha=e^{i\pi/3}$, $\beta=e^{-i\pi/3}$.

Step 2: Compute
$\alpha^{200}=e^{i200\pi/3}=e^{i(66\pi+2\pi/3)}=e^{2i\pi/3}$ and $\beta^{206}=e^{-i206\pi/3}=e^{-i(68\pi+2\pi/3)}=e^{-2i\pi/3}$.
Their sum is $2\cos(2\pi/3) = -1$, so the total is $-1+2=1$, option (A).

Final Answer:
The expression equals 1, option (A). \[ \boxed{1} \]