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Answer 1

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$\displaystyle\left|\begin{array}{ccc}\alpha&\beta&\gamma\\ \alpha^2&\beta^2&\gamma^2\\ \beta+\gamma&\gamma+\alpha&\alpha+\beta\end{array}\right| \\ \\= \left|\begin{array}{ccc}\alpha&\beta-\gamma&\gamma\\ \alpha^2&\beta^2-\gamma^2&\gamma^2\\ \beta+\gamma&(\gamma+\alpha)-(\alpha+\beta)&\alpha+\beta\end{array}\right| \\ \\= (\beta-\gamma) \left|\begin{array}{ccc}\alpha&1&\gamma\\ \alpha^2&\beta+\gamma&\gamma^2\\ \beta+\gamma&-1&\alpha+\beta\end{array}\right|$

$= (\beta-\gamma) \left|\begin{array}{ccc}\alpha&1&\gamma-\alpha\\ \alpha^2&\beta+\gamma&\gamma^2-\alpha^2\\ \beta+\gamma&-1&(\alpha+\beta)-(\beta+\gamma))\end{array}\right| \\ \\= (\beta-\gamma)(\gamma-\alpha) \left|\begin{array}{ccc}\alpha&1&1\\ \alpha^2&\beta+\gamma&\gamma+\alpha\\ \beta+\gamma&-1&-1\end{array}\right| \\ \\= (\beta-\gamma)(\gamma-\alpha) \left|\begin{array}{ccc}\alpha&1&0\\ \alpha^2&\beta+\gamma&\alpha-\beta\\ \beta+\gamma&-1&0\end{array}\right|$

$= (\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)\times - \left|\begin{array}{cc}\alpha&1\\ \beta+\gamma&-1\end{array}\right| \\ \\= (\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$