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The value of the determinant $\left|\begin{array}{ccc}1-\alpha & \alpha-\alpha^{2} & \alpha^{2} \\ 1-\beta & \beta-\beta^{2} & \beta^{2} \\ 1-\gamma & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$ is equal to
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The correct answer is:
$(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
$\left|\begin{array}{lll}1-\alpha & \alpha-\alpha^{2} & \alpha^{2} \\ 1-\beta & \beta-\beta^{2} & \beta^{2} \\ 1-\gamma & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$
$\mathrm{c}_{1} \rightarrow \mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha-\alpha^{2} & \alpha^{2} \\ 1 & \beta-\beta^{2} & \beta^{2} \\ 1 & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$
$\mathrm{c}_{2} \rightarrow \mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha & \alpha^{2} \\ 1 & \beta & \beta^{2} \\ 1 & \gamma & \gamma^{2}\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
$\mathrm{c}_{1} \rightarrow \mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha-\alpha^{2} & \alpha^{2} \\ 1 & \beta-\beta^{2} & \beta^{2} \\ 1 & \gamma-\gamma^{2} & \gamma^{2}\end{array}\right|$
$\mathrm{c}_{2} \rightarrow \mathrm{c}_{2}+\mathrm{c}_{3}$
$=\left|\begin{array}{lll}1 & \alpha & \alpha^{2} \\ 1 & \beta & \beta^{2} \\ 1 & \gamma & \gamma^{2}\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$
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