If , then find values of x.

Given, $\quad\left|\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0$

[Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ], we have

$\Rightarrow \quad\left|\begin{array}{ccc}12+x & 12+x & 12+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right|=0$

Now,

$\left[\right.$ Taking $(12+x)$ common from $\left.R_{1}\right]$

$\Rightarrow \quad(12+x)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & -2 x & 4+x \\ 2 x & 2 x & 4-x\end{array}\right|=0$

$\Rightarrow \quad(12+x)(0-(-2 x)(2 x)]=0$

$(12+x)\left(4 x^{2}\right)=0$

Hence, $\quad x=-12,0$