If `|[4-x, 4+x, 4+x], [4+x, 4-x, 4+x],[4+x, 4+x, 4-x]| = 0` find the value of x.

Explanation

To solve the determinant equation \(|[4-x, 4+x, 4+x], [4+x, 4-x, 4+x], [4+x, 4+x, 4-x]| = 0\), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} \]

### Step 2: Apply Row Transformation
We can simplify the determinant by transforming the first row \(R_1\) to \(R_1 + R_2 + R_3\):
\[ R_1 \rightarrow R_1 + R_2 + R_3 \]
This gives us:
\[ D = \begin{vmatrix} (4-x) + (4+x) + (4+x) & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} \]
Calculating the first row:
\[ = \begin{vmatrix} 12 & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} \]

### Step 3: Factor Out Common Terms
Next, we can factor out \(12 + x\) from the first column:
\[ D = (12 + x) \begin{vmatrix} 1 & \frac{4+x}{12} & \frac{4+x}{12} \\ \frac{4+x}{12} & \frac{4-x}{12} & \frac{4+x}{12} \\ \frac{4+x}{12} & \frac{4+x}{12} & \frac{4-x}{12} \end{vmatrix} \]

### Step 4: Column Transformation
Now, we perform column transformations \(C_1 \rightarrow C_1 - C_2\) and \(C_2 \rightarrow C_2 - C_3\):
\[ D = (12 + x) \begin{vmatrix} 1 - 1 & 0 & 0 \\ \frac{4+x}{12} - \frac{4-x}{12} & \frac{4-x}{12} - \frac{4+x}{12} & \frac{4+x}{12} \\ \frac{4+x}{12} - \frac{4+x}{12} & \frac{4+x}{12} - \frac{4-x}{12} & \frac{4-x}{12} \end{vmatrix} \]
This simplifies to:
\[ D = (12 + x) \begin{vmatrix} 0 & 0 & 0 \\ \frac{2x}{12} & -\frac{2x}{12} & \frac{4+x}{12} \\ 0 & \frac{2x}{12} & \frac{4-x}{12} \end{vmatrix} \]

### Step 5: Expand the Determinant
Since the first column has all zeros, we can expand the determinant:
\[ D = (12 + x) \cdot 0 = 0 \]
Thus, we need to consider the factors:
1. \(12 + x = 0\)
2. The determinant of the remaining matrix must also equal zero.

### Step 6: Solve for \(x\)
From \(12 + x = 0\):
\[ x = -12 \]

### Step 7: Solve the Quadratic
Now, we also need to solve the quadratic equation from the determinant:
\[ 4x^2 = 0 \implies x = 0 \]

### Final Values
The possible values of \(x\) are:
\[ x = 0 \quad \text{and} \quad x = -12 \]

### Summary of Solutions
Thus, the final answer is:
\[ x = 0 \quad \text{or} \quad x = -12 \]