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If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-3 x^2+x+5=0$, then $y=\Sigma \alpha^2+\alpha \beta \gamma$ satisfies the equation
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Verified Answer
The correct answer is:
$y^3-y^2-y-2=0$
Given, $\alpha, \beta$ and $\gamma$ are roots of
$x^3-3 x^2+x+5=0$
$\therefore \quad \alpha+\beta+\gamma=3$
$\alpha \beta+\beta \gamma+\gamma \alpha=1$
$\alpha \beta \gamma=-5$
Now, $y=\Sigma \alpha^2+\alpha \beta \gamma$
$\begin{aligned} & y=\left(\alpha^2+\beta^2+\gamma^2\right)+\alpha \beta \gamma \\ & y=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)+\alpha \beta \gamma \\ & y=(3)^2-2(1)-5=9-7=2\end{aligned}$
$\therefore \quad y=2$
$y=2$ only satisfied the equation
$y^3-y^2-y-2=0$
$x^3-3 x^2+x+5=0$
$\therefore \quad \alpha+\beta+\gamma=3$
$\alpha \beta+\beta \gamma+\gamma \alpha=1$
$\alpha \beta \gamma=-5$
Now, $y=\Sigma \alpha^2+\alpha \beta \gamma$
$\begin{aligned} & y=\left(\alpha^2+\beta^2+\gamma^2\right)+\alpha \beta \gamma \\ & y=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)+\alpha \beta \gamma \\ & y=(3)^2-2(1)-5=9-7=2\end{aligned}$
$\therefore \quad y=2$
$y=2$ only satisfied the equation
$y^3-y^2-y-2=0$
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