If alpha, beta, gamma are the roots of the eq | vedclass

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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 find the value of $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2$. $\Sigma \alpha^2 \beta+\Sigma \alpha \beta^2 = \alpha^2 \beta+\beta^2 \gamma+\gamma^2 \alpha+\alpha \beta^2+\beta \gamma^2+\gamma \alpha^2$ $= \alpha \beta(\alpha+\beta)+\beta \gamma(\beta+\gamma)+\gamma \alpha(\gamma+\alpha)$ $= \alpha \beta(6-\gamma)+\beta \gamma(6-\alpha)+\gamma \alpha(6-\beta)$ $= 6(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma$ $= 6(11)-3(-6)$ $= 66+18 = 84$.

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

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