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(B) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x+6=0$. From the relation between roots and coefficients: $\alpha+\be

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(B) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x+6=0$. From the relation between roots and coefficients: $\alpha+\beta+\gamma = 6$ $\alpha \beta+\beta \gamma+\gamma \alpha = 11$ $\alpha \beta \gamma = -6$ We need to evaluate $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$. Note that $(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha) = (\alpha^2 \beta+\alpha \beta \gamma+\alpha^2 \gamma) + (\alpha \beta^2+\beta^2 \gamma+\alpha \beta \gamma) + (\alpha \beta \gamma+\beta \gamma^2+\gamma^2 \alpha)$ $= (\alpha^2 \beta+\alpha \beta^2+\beta^2 \gamma+\beta \gamma^2+\gamma^2 \alpha+\gamma \alpha^2) + 3 \alpha \beta \gamma$ $= (\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2) + 3 \alpha \beta \gamma$ Therefore,$\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2 = (\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha) - 3 \alpha \beta \gamma$ $= (6)(11) - 3(-6)$ $= 66 + 18 = 84$.

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

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