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If $\alpha, \beta, \gamma$ are the roots of $2 x^3-2 x-1=0$, then $(\Sigma \alpha \beta)^2$ is equal to
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The correct answer is:
$1$
$\alpha, \beta$ and $\gamma$ are the roots of the equation
$$
\begin{array}{rlrl}
2 x^3-2 x-1 & =0 \\
\text { or } & 2 x^3+0 x^2-2 x-1 & =0 \\
\Rightarrow \quad & \alpha \beta+\beta \gamma+\gamma \alpha=\frac{(-2)}{2} & =-1 \\
\text { Now, }(\Sigma \alpha \beta)^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 & =(-1)^2=1
\end{array}
$$
$$
\begin{array}{rlrl}
2 x^3-2 x-1 & =0 \\
\text { or } & 2 x^3+0 x^2-2 x-1 & =0 \\
\Rightarrow \quad & \alpha \beta+\beta \gamma+\gamma \alpha=\frac{(-2)}{2} & =-1 \\
\text { Now, }(\Sigma \alpha \beta)^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 & =(-1)^2=1
\end{array}
$$
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