Thus, $ (x^2 + x + 1)^3 = 0 $ at $ x = \omega $, so $ x^2 + x + 1 $ is divisible by $ x^2 - x + 1 $, since both are degree 2 and the first vanishes where the second does.

February 22, 2026

$Thus, $ (x^2 + x + 1)^3 = 0 $ at $ x = \omega $, so $ x^2 + x + 1 $ is divisible by $ x^2 - x + 1 $, since both are degree 2 and the first vanishes where the second does.$

e 1 $, and $ \omega^2 + \omega + 1 = 0 $.
We evaluate $ (x^2 + x + 1)^3 $ at $ x = \omega $:
\omega^2 + \omega + 1 = 0 \Rightarrow x^2 + x + 1 = 0 ext{ at } x = \omega
Therefore, the remainder is 0.
Hence, the final answer is:
$$### Question 1

Thus, $ (x^2 + x + 1)^3 = 0 $ at $ x = \omega $, so $ x^2 + x + 1 $ is divisible by $ x^2 - x + 1 $, since both are degree 2 and the first vanishes where the second does.

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