Sometimes it is easier for us to evaluate a determinant if we can express it as product of two determinant. For instance, if \( S_{k}=a^{k}+b^{k}+c^{k} \), then
\( \left|\begin{array}{lll}S_{0} & S_{1} & S_{2} \\ S_{1} & S_{2} & S_{3} \\ S_{2} & S_{3} & S_{4}\end{array}\right| \) can be written as \( \Delta^{2} \)
where \( \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right| \)
which is quite easy to evaluate.
If \( a_{1}a_{2}a_{3}, b_{1}b_{2}b_{3} \) and \( a_{i} b_{j} \neq 1 \) for \( 1 \leq i, j \) \( \leq 3 \), then the determinant
\[
\Delta=\left|\begin{array}{ccc}
\frac{1-a_{1}^{3} b_{1}^{3}}{1-a_{1} b_{1}} & \frac{1-a_{1}^{3} b_{2}^{3}}{1-a_{1} b_{2}} & \frac{1-a_{1}^{3} b_{3}^{3}}{1-a_{1} b_{3}} \\
\frac{1-a_{2}^{3} b_{1}^{3}}{1-a_{2} b_{1}} & \frac{1-a_{2}^{3} b_{2}^{3}}{1-a_{2} b_{2}} & \frac{1-a_{2}^{3} b_{3}^{3}}{1-a_{2} b_{3}} \\
\frac{1-a_{3}^{3} b_{1}^{3}}{1-a_{3} b_{1}} & \frac{1-a_{3}^{3} b_{2}^{3}}{1-a_{3} b_{2}} & \frac{1-a_{3}^{3} b_{3}^{3}}{1-a_{3} b_{3}}
\end{array}\right|
\]
is
(a) positive
(b) non-negative
(c) negative
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