lf $\alpha,\ \beta$ are the roots of the equation $x^{2}+x+1=0$ and $S_k={\alpha}^k+{\beta}^k ; k=1,2,3,4 $ , then $\left| \begin{matrix} 3 & 1+{ S }{ 1 } & 1+{ S }{ 2 } \ 1+{ S }{ 1 } & 1+{ S }{ 2 } & 1+{ S }{ 3 } \ 1+{ S }{ 2 } & 1+{ S }{ 3 } & 1+{ S }{ 4 } \end{matrix} \right| =\ $

Since alpha and beta are the roots of x squared plus x plus 1 equals 0, the sum of the roots S1 is negative 1. Using the recurrence relation from the equation (x squared equals negative x minus 1), we find S2 equals 1 and S3 equals negative 2. Substituting these into the determinant gives the matrix 3, 0, 2 in the first row, 0, 2, negative 1 in the second row, and 2, negative 1, negative 2 in the third row. Evaluating this determinant using expansion yields negative 27.

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