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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Aicharya · Accepted Answer

$\alpha, \beta$ and $\gamma$ are the roots of the equation $ \begin{array}{rlrl} 2 x^3-2 x-1 & =0 \ 	ext { or } & 2 x^3+0 x^2-2 x-1 & =0 \ \Rightarrow \quad & \alpha \beta+\beta \gamma+\gamma \alpha=\frac{(-2)}{2} & =-1 \ 	ext { Now, }(\Sigma \alpha \beta)^2=(\alpha \beta+\beta \gamma+\gamma \alpha)^2 & =(-1)^2=1 \end{array} $

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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