Using properties of determinants. Find the value of 'x'
`|(4-x,4+x,4+x),(4+x,4-x,4+x),(4-x,4+x,4+x)|=0`

Answer

To solve the determinant equation

\[ \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4-x & 4+x & 4+x \end{vmatrix} = 0, \]

we will use properties of determinants.

### 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 operations

We can simplify the determinant by performing row operations. Let's subtract the first row from the second and the third rows:

\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1. \]

This gives us:

\[ D = \begin{vmatrix} 4-x & 4+x & 4+x \\ (4+x) - (4-x) & (4-x) - (4+x) & (4+x) - (4+x) \\ (4-x) - (4-x) & (4+x) - (4+x) & (4+x) - (4+x) \end{vmatrix}. \]

Calculating the new rows:

- For \(R_2\):
- First element: \((4+x) - (4-x) = 2x\)
- Second element: \((4-x) - (4+x) = -2\)
- Third element: \(0\)

- For \(R_3\):
- All elements become \(0\).

So, the determinant simplifies to:

\[ D = \begin{vmatrix} 4-x & 4+x & 4+x \\ 2x & -2 & 0 \\ 0 & 0 & 0 \end{vmatrix}. \]

### Step 3: Evaluate the determinant

Since the third row consists entirely of zeros, the determinant \(D\) is equal to zero:

\[ D = 0. \]

### Step 4: Conclusion

Since the determinant equals zero, we conclude that the original determinant is zero for all values of \(x\). Therefore, the value of \(x\) can be any real number.

### Final Answer

The value of \(x\) is:

\[ \text{For all } x \in \mathbb{R}. \]