If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6 x^2+11 x+6=0$, then $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$ is equal to :

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Aicharya · Accepted Answer

$\because \alpha, \beta, \gamma$ are the roots of the equation $x^3-6 x^2+11 x+6=0$ $	herefore \quad \alpha+\beta+\gamma=6$ $\alpha \beta+\beta \gamma+\gamma \alpha=11$ and $\quad \alpha \beta \gamma=-6$ Now $\quad \Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$ $=\alpha^2 \beta+\beta^2 \gamma+\gamma^2 \alpha+\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2$ $=\alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\gamma+\alpha)$ $=\alpha \beta(6-\gamma)+\beta \gamma(6-\alpha)+\gamma \alpha(6-\beta)$ $=6(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma$ $=6(11)+3(6)$ $=66+18=84$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6 x^2+11 x+6=0$, then $\Sigma \alpha^2…

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