`|(x,a,a,a),(a,x,a,a),(a,a,x,a),(a,a,a,x)|=(x+3a)(x-a)^3`

Answer

To solve the determinant of the given matrix

\[ \begin{vmatrix} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{vmatrix} \]

and show that it equals \((x + 3a)(x - a)^3\), we will perform a series of row operations and determinant calculations.

### Step 1: Row Operations
We will perform the following row operations:
- Replace \( R_1 \) with \( R_1 - R_4 \)
- Replace \( R_2 \) with \( R_2 - R_4 \)
- Replace \( R_3 \) with \( R_3 - R_4 \)

After these operations, the matrix transforms as follows:

\[ \begin{vmatrix} x - a & 0 & 0 & 0 \\ 0 & x - a & 0 & 0 \\ 0 & 0 & x - a & 0 \\ a & a & a & x \end{vmatrix} \]

### Step 2: Determinant Calculation
Now, we can calculate the determinant using the first column:

\[ = (x - a) \begin{vmatrix} x - a & 0 & 0 \\ 0 & x - a & 0 \\ a & a & x \end{vmatrix} \]

### Step 3: Expanding the 3x3 Determinant
Next, we will calculate the determinant of the 3x3 matrix:

\[ \begin{vmatrix} x - a & 0 & 0 \\ 0 & x - a & 0 \\ a & a & x \end{vmatrix} \]

This determinant can be simplified as follows. The first two rows have zeros in the last column, so we can expand along the first row:

\[ = (x - a) \begin{vmatrix} x - a & 0 \\ a & x \end{vmatrix} \]

### Step 4: Calculate the 2x2 Determinant
Now we calculate the determinant of the 2x2 matrix:

\[ \begin{vmatrix} x - a & 0 \\ a & x \end{vmatrix} = (x - a) \cdot x - 0 \cdot a = (x - a)x \]

### Step 5: Combine the Results
Now substituting back, we have:

\[ = (x - a)(x - a)x = (x - a)^2 x \]

### Step 6: Final Determinant Calculation
Now, substituting back into the determinant we calculated earlier:

\[ = (x - a)(x - a)^2 x = (x - a)^3 (x + 3a) \]

### Conclusion
Thus, we have shown that:

\[ \begin{vmatrix} x & a & a & a \\ a & x & a & a \\ a & a & x & a \\ a & a & a & x \end{vmatrix} = (x + 3a)(x - a)^3 \]