2023 SSMO Accuracy Round Problems/Problem 6

Problem

Let the roots of $P(x) = x^3 - 2023x^2 + 2023^{2023}$ be $\alpha, \beta, \gamma.$ . Find $\frac{\alpha^2 + \beta^2}{\alpha + \beta} + \frac{\beta^2 + \gamma^2}{\beta+\gamma} + \frac{\gamma^2 + \alpha^2}{\gamma + \alpha}$

Solution

By Vieta's relation we get, $\alpha+\beta+\gamma=2023=\sum_{cyc}{}\alpha^2=4092529.$ $\alpha\beta+\beta\gamma+\gamma\alpha=0$ $\alpha\beta\gamma=-(2023)^{2023} \implies \sum_{cyc}{}\alpha^3=8279186167-3(2023)^{2023}$ Now we have to find the value of $\frac{\alpha^2+\beta^2}{\alpha+\beta}+\frac{\beta^2+\gamma^2}{\beta+\gamma}+\frac{\gamma^2+\alpha^2}{\gamma+\alpha}=\frac{4092529-\alpha^2}{2023-\alpha}+\frac{4092529-\beta^2}{2023-\beta}+\frac{4092529-\gamma^2}{2023-\gamma}$ Therefore we get, $\frac{\sum_{cyc}{}(4092529-\alpha^2)(2023-\beta)(2023-\gamma)}{\prod_{cyc}{}(2023-\alpha)}$ We get, $\frac{-\alpha\beta\gamma(\sum_{cyc}{}\alpha)+2023\sum_{cyc}{}\alpha^2(\beta+\gamma)-4092529(\sum_{cyc}{}\alpha^2)+4092529(\sum_{cyc}{}\alpha\beta)-8279186167(2(\sum_{cyc}{}\alpha))+50246380847523}{\prod_{cyc}{}(2023-\alpha)}$ $\frac{4(2023)^{2024}}{2023^{2023}}\implies\boxed{8092.}$

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2023 SSMO Accuracy Round Problems/Problem 6

Problem

Solution