The values of x in the following determinant | vedclass

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(D) Let $\Delta = \left| \begin{array}{ccc} a+x & a-x & a-x \ a-x & a+x & a-x \ a-x & a-x & a+x \end{array} ight| = 0$. Applying $C_1 	o C_1 + C_

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$x = 0, x = 3a$

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(D) Let $\Delta = \left| \begin{array}{ccc} a+x & a-x & a-x \ a-x & a+x & a-x \ a-x & a-x & a+x \end{array} ight| = 0$. Applying $C_1 	o C_1 + C_2 + C_3$: $\Delta = \left| \begin{array}{ccc} 3a-x & a-x & a-x \ 3a-x & a+x & a-x \ 3a-x & a-x & a+x \end{array} ight| = 0$. Taking $(3a-x)$ common from $C_1$: $(3a-x) \left| \begin{array}{ccc} 1 & a-x & a-x \ 1 & a+x & a-x \ 1 & a-x & a+x \end{array} ight| = 0$. Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $(3a-x) \left| \begin{array}{ccc} 1 & a-x & a-x \ 0 & 2x & 0 \ 0 & 0 & 2x \end{array} ight| = 0$. Expanding along $C_1$: $(3a-x) [1(4x^2 - 0)] = 0$. $(3a-x)(4x^2) = 0$. Therefore,$x = 3a$ or $x^2 = 0$,which implies $x = 0$. Thus,the values are $x = 0, x = 3a$.

The values of $x$ in the following determinant equation,$\left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$ are

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