If $\begin{vmatrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{vmatrix} = 0$, then find the values of $x$.

$0, -12$

The correct answer is Option (2) → $0, -12$ ##

Given, $\begin{vmatrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{vmatrix} = 0$

On expanding along $R_1$, we get

$(4 - x) [(4 - x)^2 - (4 + x)^2] - (4 + x) [(4 + x)(4 - x) - (4 + x)^2] + (4 + x) [(4 + x)^2 - (4 + x)(4 - x)] = 0$

$\Rightarrow (4 - x) [16 + x^2 - 8x - 16 - x^2 - 8x] - (4 + x) [4 - x - 4 - x] + (4 + x)^2 [(4 + x) - 4 + x] = 0$

$\Rightarrow (4 - x) [-16x] - (4 + x)^2 [-2x] + (4 + x)^2 [2x] = 0$

$\Rightarrow (4 - x) [-16x] + (4 + x)^2 [2x + 2x] = 0$

$\Rightarrow -64x + 16x^2 + 64x + 4x^3 + 32x^2 = 0$

$\Rightarrow 4x^3 + 48x^2 = 0$

$\Rightarrow 4x^2(x + 12) = 0 \Rightarrow x^2 = 0 \text{ or } x + 12 = 0$

$\Rightarrow x = 0 \text{ or } x = -12$

Hence, the values of $x$ are $0$ and $-12$.