This is in response to Soner Karaca, who asked me the following questions. Everyone is free to provide feedback.

**Problem:**

- Given \(a^2+b^2+ab-\sqrt 3b+1=0\), find \(b-a\).
- Find \(7^{77} \text{ mod } 1000\).

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## Comments

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TopNewestSolution to Problem 1\(\begin{align} a^2+b^2+ab-\sqrt 3b+1 & =0 \\ b^2 + (a-\sqrt 3)b + a^2+ 1 & = 0 \\ \implies b & = \frac {\sqrt 3-a \pm \sqrt{a^2-2\sqrt 3a + 3 - 4a^2-4}}2 \\ & = \frac {\sqrt 3-a \pm \sqrt{-3a^2-2\sqrt 3a -1}}2 \\ & = \frac {\sqrt 3-a \pm (\sqrt3 a+1)i}2 \\ \implies b - a & = \frac {\sqrt 3-3a \pm (\sqrt3 a+1)i}2 \end{align} \)

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sir, what is this note about? I mean is this note written to tell the solution for for some other purpose.. Although I have understood the solution you wrote of the above problem.

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I was replying to @Soner Karaca who asked me this problem through Messenger. I need LaTex to reply him.

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ok sir.

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