If a + b + c = 0,then the solution of the equ | vedclass

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(C) Given the determinant equation: $\left| \begin{array}{ccc} a - x & c & b \ c & b - x & a \ b & a & c - x \end{array} ight| = 0$.Apply the oper

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$0, \pm \sqrt{\frac{3}{2}(a^2 + b^2 + c^2)}$

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(C) Given the determinant equation: $\left| \begin{array}{ccc} a - x & c & b \ c & b - x & a \ b & a & c - x \end{array} ight| = 0$.Apply the operation $C_1 	o C_1 + C_2 + C_3$:$\left| \begin{array}{ccc} a + b + c - x & c & b \ a + b + c - x & b - x & a \ a + b + c - x & a & c - x \end{array} ight| = 0$.Since $a + b + c = 0$,the first column becomes $-x$ in every entry:$\left| \begin{array}{ccc} -x & c & b \ -x & b - x & a \ -x & a & c - x \end{array} ight| = 0$.Taking $-x$ common from the first column:$-x \left| \begin{array}{ccc} 1 & c & b \ 1 & b - x & a \ 1 & a & c - x \end{array} ight| = 0$.This gives $x = 0$ as one solution.For the remaining part,expand the determinant:$1((b-x)(c-x) - a^2) - c(c-x-a) + b(a-b+x) = 0$.Simplifying this leads to $x^2 - (a^2 + b^2 + c^2) + (ab + bc + ca) = 0$.Since $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$,we have $ab+bc+ca = -\frac{1}{2}(a^2+b^2+c^2)$.Substituting this: $x^2 - (a^2+b^2+c^2) - \frac{1}{2}(a^2+b^2+c^2) = 0$.$x^2 = \frac{3}{2}(a^2+b^2+c^2) \Rightarrow x = \pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$.Thus,the solutions are $x = 0, \pm \sqrt{\frac{3}{2}(a^2+b^2+c^2)}$.

If $a + b + c = 0$,then the solution of the equation $\left| \begin{array}{ccc} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{array} \right| = 0$ is

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