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(A) Let the $n^{	ext{th}}$ roots of unity be $z_0, z_1, \ldots, z_{n-1}$ where $z_0 = 1$. The equation $x^n - 1 = 0$ has roots $1, \alpha_1, \alpha_2

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(A) Let the $n^{	ext{th}}$ roots of unity be $z_0, z_1, \ldots, z_{n-1}$ where $z_0 = 1$. The equation $x^n - 1 = 0$ has roots $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$.By Vieta's formulas,the sum of the roots taken two at a time is the coefficient of $x^{n-2}$ divided by the coefficient of $x^n$.For $x^n - 1 = 0$,the coefficient of $x^{n-1}$ is $0$ and the coefficient of $x^{n-2}$ is $0$ (for $n > 2$).Let $S = \sum_{0 \leq i This sum can be expanded as: $z_0(\sum_{i=1}^{n-1} \alpha_i) + \sum_{1 \leq i Since $z_0 = 1$ and the sum of all $n^{	ext{th}}$ roots of unity is $0$,we have $1 + \sum_{i=1}^{n-1} \alpha_i = 0$,so $\sum_{i=1}^{n-1} \alpha_i = -1$.Substituting this into the expansion: $1(-1) + \sum_{1 \leq i Therefore,$\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j = 1$.

If $1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}$ are the $n^{\text{th}}$ roots of unity,then $\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j =$

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