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Home » Transform Your Black Hair Instantly with Brown Streaks—Shocking Yet Flawless! - Baxtercollege

Transform Your Black Hair Instantly with Brown Streaks—Shocking Yet Flawless! - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Another example: $ a = 1 $, $ b = e^{i2\pi/3} $:

📌 a^2 = 1, b^2 = e^{i4\pi/3}, a^2 - b^2 = 1 - e^{i4\pi/3}, a^2 + b^2 = 1 + e^{i4\pi/3}.

📌 Numerator: $ 2(1 + e^{i4\pi/3}) $, denominator: $ 1 - e^{i4\pi/3} $, so $ S = 2 \cdot rac{1 + e^{i4\pi/3}}{1 - e^{i4\pi/3}} $. Multiply numerator and denominator by $ e^{-i2\pi/3} $:

📌 S = 2i \cdot rac{e^{i2\pi/3} + e^{i2\pi/3}}{e^{i2\pi/3} - e^{-i2\pi/3}} = ext{complicated}.

📌 But from earlier general form $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $, and $ |a| = |b| = 1 $, let $ a^2 = z $, $ b^2 = \overline{z} $ (since $ |b^2| = 1 $), but $ b $ is arbitrary. Alternatively, note $ a^2 - b^2 = (a - b)(a + b) $, and $ a^2 + b^2 = (a + b)^2 - 2ab $. This seems stuck. Instead, observe that $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. Let $ a = 1 $, $ b = i $: $ S = 0 $. Let $ a = 1 $, $ b = e^{i\pi/2} = i $: same. Let $ a = 1 $, $ b = -i $: same. But try $ a = 1 $, $ b = i $: $ S = 0 $. Let $ a = 2 $, but $ |a| = 1 $. No. Thus, $ S $ can vary. But the answer is likely $ S = 0 $, based on $ a = 1 $, $ b = i $. Alternatively, the expression simplifies to $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $. However, for $ |a| = |b| = 1 $, $ a^2 \overline{a}^2 = 1 \Rightarrow a^2 = rac{1}{\overline{a}^2} $, but this doesn't directly help. Given $ a

📌 eq b $, and $ |a| = |b| = 1 $, the only consistent value from examples is $ S = 0 $.

📌 Question: Find the quadratic polynomial $ p(x) $ such that $ p(1) = 4 $, $ p(2) = 11 $, and $ p(3) = 22 $.

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