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If $\alpha, \beta$ and $\gamma$ are the root 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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The correct answer is:
84
$\alpha, \beta$ and $\gamma$ are roots of $x^3-6 x^2+11 x+6=0$
Then, $\alpha+\beta+\gamma=6, \alpha \beta+\beta \gamma+\gamma \alpha=11$
and $\alpha \beta \gamma=-6$
Now,
$$
\begin{aligned}
\Sigma \alpha^2 \beta+ & \Sigma \alpha \beta^2 \\
& =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma \\
& =(6)(11)-3(-6)=66+18=84
\end{aligned}
$$
Then, $\alpha+\beta+\gamma=6, \alpha \beta+\beta \gamma+\gamma \alpha=11$
and $\alpha \beta \gamma=-6$
Now,
$$
\begin{aligned}
\Sigma \alpha^2 \beta+ & \Sigma \alpha \beta^2 \\
& =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma \\
& =(6)(11)-3(-6)=66+18=84
\end{aligned}
$$
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