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(B) Given the cubic equation $2x^3 + 0x^2 - 2x - 1 = 0$.Let the roots be $\alpha, \beta, \gamma$.According to Vieta's formulas for a cubic equation $a

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$1$

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(B) Given the cubic equation $2x^3 + 0x^2 - 2x - 1 = 0$.Let the roots be $\alpha, \beta, \gamma$.According to Vieta's formulas for a cubic equation $ax^3 + bx^2 + cx + d = 0$,the sum of the roots taken two at a time is given by $\Sigma \alpha \beta = \frac{c}{a}$.Here,$a = 2$,$b = 0$,$c = -2$,and $d = -1$.Therefore,$\Sigma \alpha \beta = \frac{-2}{2} = -1$.We need to find $(\Sigma \alpha \beta)^2$.$(\Sigma \alpha \beta)^2 = (-1)^2 = 1$.

If $\alpha, \beta, \gamma$ are the roots of $2x^3 - 2x - 1 = 0$,then $(\Sigma \alpha \beta)^2$ is equal to

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