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(B) Let $\Delta = \left| {\begin{array}{ccc} a + b & b + c & c + a \ b + c & c + a & a + b \ c + a & a + b & b + c \end{array}} ight|$.Applying $R

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$2$

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(B) Let $\Delta = \left| {\begin{array}{ccc} a + b & b + c & c + a \ b + c & c + a & a + b \ c + a & a + b & b + c \end{array}} ight|$.Applying $R_1 	o R_1 + R_2 + R_3$,we get:$\Delta = \left| {\begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \ b + c & c + a & a + b \ c + a & a + b & b + c \end{array}} ight| = 2(a+b+c) \left| {\begin{array}{ccc} 1 & 1 & 1 \ b + c & c + a & a + b \ c + a & a + b & b + c \end{array}} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = 2(a+b+c) \left| {\begin{array}{ccc} 1 & 0 & 0 \ b + c & a - b & a - c \ c + a & b - c & b - a \end{array}} ight| = 2(a+b+c) [-(a-b)^2 - (a-c)(b-c)] = -2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.This simplifies to $-2(a^3+b^3+c^3-3abc)$.Now,the determinant $D = \left| {\begin{array}{ccc} a & b & c \ b & c & a \ c & a & b \end{array}} ight| = a(bc-a^2) - b(b^2-ac) + c(ab-c^2) = -(a^3+b^3+c^3-3abc)$.Comparing $\Delta = K \cdot D$,we get $-2(a^3+b^3+c^3-3abc) = K \cdot -(a^3+b^3+c^3-3abc)$,so $K = 2$.

$\left| {\begin{array}{ccc} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right| = K \left| {\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right|$,then $K = $

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