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Home » Get Easy Credit Cards Fast—Illuminating the Top Options You Need! - ClickBalance

Get Easy Credit Cards Fast—Illuminating the Top Options You Need! - ClickBalance

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 Let \( w = z^3 \). Then the equation becomes:

📌 w^2 + w + 1 = 0

📌 w = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}

📌 So \( z^3 = \omega \) or \( z^3 = \omega^2 \), where \( \omega = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \), and \( \omega^2 = e^{-2\pi i/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \), both primitive cube roots of unity.

📌 We now solve \( z^3 = \omega \) and \( z^3 = \omega^2 \), each giving 3 roots.

📌 Let’s find the roots of \( z^3 = \omega \). Write \( \omega = e^{2\pi i/3} \), so:

📌 z = \omega^{1/3} \cdot e^{2\pi i k / 3} = e^{2\pi i/9} \cdot e^{2\pi i k / 3} = e^{2\pi i (1/9 + k/3)} = e^{2\pi i (1/9 + 3k/9)} = e^{2\pi i (3k+1)/9}, \quad k = 0,1,2

📌 Similarly, for \( z^3 = \omega^2 = e^{-2\pi i/3} \):

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