If the complex cube roots of (-i) are alpha, | vedclass

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(D) Let the roots be $\alpha, \beta, \gamma$. These are the roots of the equation $z^3 = -i$,or $z^3 + i = 0$.By Vieta's formulas,for a cubic equation

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(D) Let the roots be $\alpha, \beta, \gamma$. These are the roots of the equation $z^3 = -i$,or $z^3 + i = 0$.By Vieta's formulas,for a cubic equation $az^3 + bz^2 + cz + d = 0$,the sum of the roots $\alpha + \beta + \gamma = -b/a$ and the sum of the products of roots taken two at a time $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$.Here,$a = 1$,$b = 0$,$c = 0$,and $d = i$.Thus,$\alpha + \beta + \gamma = 0$ and $\alpha\beta + \beta\gamma + \gamma\alpha = 0$.We know the identity $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$.Substituting the values,we get $\alpha^2 + \beta^2 + \gamma^2 = (0)^2 - 2(0) = 0$.

If the complex cube roots of $(-i)$ are $\alpha, \beta, \gamma$,then $\alpha^2+\beta^2+\gamma^2=$

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