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(A) Let $\Delta = \left| \begin{array}{ccc} a & b & c \ b & c & a \ c & a & b \end{array} ight|$.Expanding the determinant,we get:$\Delta = a(cb -

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Negative

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(A) Let $\Delta = \left| \begin{array}{ccc} a & b & c \ b & c & a \ c & a & b \end{array} ight|$.Expanding the determinant,we get:$\Delta = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)$$= abc - a^3 - b^3 + abc + abc - c^3$$= 3abc - (a^3 + b^3 + c^3)$$= -(a^3 + b^3 + c^3 - 3abc)$.Using the algebraic identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$,we have:$\Delta = -(a + b + c) 	imes \frac{1}{2} [(a - b)^2 + (b - c)^2 + (c - a)^2]$.Since $a, b, c$ are positive,$(a + b + c) > 0$. Since $a, b, c$ are not all equal,at least one of $(a - b)^2, (b - c)^2, (c - a)^2$ must be positive,so the sum of squares is positive.Therefore,$\Delta$ is negative.

If $a, b, c$ are positive and not all equal,then the value of the determinant $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right|$ is

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