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If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-8 x+8=0$, then $\Sigma \alpha^{2}$ and $\Sigma \frac{1}{\alpha \beta}$ are respectively is equal to
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16 and 0
Since, $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^{3}-8 x+8=0$, then
$\begin{gathered}
\alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=-8 \\
\alpha \beta \gamma=-8 ...(i)
\end{gathered}$
Therefore, $(\alpha+\beta+\gamma)^{2}=0$
$\Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \beta+\beta \gamma+\gamma \alpha)=0$
$\begin{aligned} \Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2} &=-2(-8)=16 \\ \Sigma \alpha^{2} &=16 \end{aligned}$
and
$\begin{gathered}
\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{0}{-8}=0 \\
\Sigma \frac{1}{\alpha \beta}=0
\end{gathered}$
$\begin{gathered}
\alpha+\beta+\gamma=0, \alpha \beta+\beta \gamma+\gamma \alpha=-8 \\
\alpha \beta \gamma=-8 ...(i)
\end{gathered}$
Therefore, $(\alpha+\beta+\gamma)^{2}=0$
$\Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}+2(\alpha \beta+\beta \gamma+\gamma \alpha)=0$
$\begin{aligned} \Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2} &=-2(-8)=16 \\ \Sigma \alpha^{2} &=16 \end{aligned}$
and
$\begin{gathered}
\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{0}{-8}=0 \\
\Sigma \frac{1}{\alpha \beta}=0
\end{gathered}$
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