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Home » So no integer. Answer is \( \sqrt[3]18 \), but recheck algebra. - Midis

So no integer. Answer is \( \sqrt[3]18 \), but recheck algebra. - Midis

📅 March 12, 2026 👤 scraface

Mar 12, 2026

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oxed{x^6 - 2x^3 + 3}
Question: Find the remainder when $ (x^2 + x + 1)^3 $ is divided by $...
Solution: Let us compute $ (x^2 + x + 1)^3 \mod (x^2 - x + 1) $.
Note that $ x^2 + x + 1 $ and $ x^2 - x + 1 $ are related via complex...
Let $ \omega $ be a primitive cube root of unity, satisfying $ \omega^...
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 = \...
Thus, $ (x^2 + x + 1)^3 = 0 $ at $ x = \omega $, so $ x^2 + x + 1 $ is...
Therefore, the remainder is 0.

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