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(D) Given the determinant equation: $\left| \begin{array}{ccc} x+a & b & c \ b & x+c & a \ c & a & x+b \end{array} ight| = 0$ Apply the column ope

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$ -(a+b+c) $

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x+a & b & c \ b & x+c & a \ c & a & x+b \end{array} ight| = 0$ Apply the column operation $C_1 	o C_1 + C_2 + C_3$: $\left| \begin{array}{ccc} x+a+b+c & b & c \ x+a+b+c & x+c & a \ x+a+b+c & a & x+b \end{array} ight| = 0$ Taking $(x+a+b+c)$ as a common factor from $C_1$: $(x+a+b+c) \left| \begin{array}{ccc} 1 & b & c \ 1 & x+c & a \ 1 & a & x+b \end{array} ight| = 0$ Setting the factor equal to zero gives: $x+a+b+c = 0$ $x = -(a+b+c)$ Thus,$-(a+b+c)$ is one of the roots of the equation.

One of the roots of the given equation $\left| \begin{array}{ccc} x+a & b & c \\ b & x+c & a \\ c & a & x+b \end{array} \right| = 0$ is

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