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(C) The determinant is given by $\Delta = \left| \begin{array}{ccc} a & b & c \ b & c & a \ c & a & b \end{array} ight|$.Expanding along the first

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$-1$

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(C) The determinant is given by $\Delta = \left| \begin{array}{ccc} a & b & c \ b & c & a \ c & a & b \end{array} ight|$.Expanding along the first row:$\Delta = a(cb - a^2) - b(b^2 - ca) + c(ab - 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)$.We know the algebraic identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.Substituting this into our expression:$\Delta = -1 	imes (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$.Comparing this with the given form $k(a + b + c)(a^2 + b^2 + c^2 - bc - ca - ab)$,we get $k = -1$.

If $\left| \begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array} \right| = k(a + b + c)(a^2 + b^2 + c^2 - bc - ca - ab)$,then $k =$

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