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(NONE) Given the equation $x^3-6x^2+11x-6=0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 6$
$\

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(NONE) Given the equation $x^3-6x^2+11x-6=0$,the roots are $\alpha, \beta, \gamma$.
From Vieta's formulas,we have:
$\alpha+\beta+\gamma = 6$
$\alpha\beta+\beta\gamma+\gamma\alpha = 11$
$\alpha\beta\gamma = 6$
We need to evaluate $\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2$.
Note that $(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) = (\alpha^2\beta + \alpha\beta\gamma + \alpha^2\gamma) + (\alpha\beta^2 + \beta^2\gamma + \alpha\beta\gamma) + (\alpha\beta\gamma + \beta\gamma^2 + \gamma^2\alpha) = \Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 + 3\alpha\beta\gamma$.
Therefore,$\Sigma \alpha^2 \beta + \Sigma \alpha \beta^2 = (\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha) - 3\alpha\beta\gamma$.
Substituting the values:
$= (6)(11) - 3(6) = 66 - 18 = 48$.

If $\alpha, \beta$ and $\gamma$ 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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