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

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $\alpha + \beta + \gamma = 6,\ \ \ \ \ \alpha\beta + \beta\gamma + \gamma\alpha = 11$
And $\alpha\beta\gamma = - 6$
Now, $\sum_{}^{}{\alpha^{2}\beta + \sum_{}^{}{\alpha\beta^{2} = \alpha^{2}\beta + \beta^{2}\alpha + \gamma^{2}\alpha + \alpha\beta^{2} + \beta\gamma^{2} + \gamma\alpha^{2}}}$
$= \alpha\beta(\alpha + \beta) + \beta\gamma(\beta + \gamma) + \gamma\alpha(\gamma + \alpha)$
$= \alpha\beta(6 - \gamma) + \beta\gamma(6 - \alpha) + \gamma\alpha(6 - \beta)$
$= 6(\alpha\beta + \beta\gamma + \gamma\alpha) - 3\alpha\beta\gamma$
$= 6(11) + 3(6) = 84$

Q. If $\alpha,\ \beta,\ \gamma$ are the roots of the equation $x^{3} - 6x^{2} + 11x + 6 = 0$, then ${\sum_{}^{}{\alpha^{2}\beta + \sum_{}^{}{\alpha\beta}}}^{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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