Show that: `|x p q p x q q q x|=(x-p)(x^2+p x-2q^2)` .

Explanation

To show that the determinant

\[ D = \begin{vmatrix} x & p & q \\ p & x & q \\ q & q & x \end{vmatrix} \]

is equal to \((x - p)(x^2 + px - 2q^2)\), we will follow these steps:

### Step 1: Set up the determinant
We start with the determinant \(D\):

\[ D = \begin{vmatrix} x & p & q \\ p & x & q \\ q & q & x \end{vmatrix} \]

### Step 2: Transform the first column
We will perform the operation \(C_1 \leftarrow C_1 - C_2\):

\[ D = \begin{vmatrix} x - p & p & q \\ p - x & x & q \\ q - q & q & x \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} x - p & p & q \\ p - x & x & q \\ 0 & q & x \end{vmatrix} \]

### Step 3: Factor out \(x - p\)
Notice that we can factor out \((x - p)\) from the first column:

\[ D = (x - p) \begin{vmatrix} 1 & p & q \\ \frac{p - x}{x - p} & x & q \\ 0 & q & x \end{vmatrix} \]

### Step 4: Simplify the second row
Now we can simplify the second row further by transforming it. We will perform the operation \(R_2 \leftarrow R_2 + R_1\):

\[ D = (x - p) \begin{vmatrix} 1 & p & q \\ 0 & x + p & q \\ 0 & q & x \end{vmatrix} \]

### Step 5: Expand the determinant
Now we can expand this determinant along the first column:

\[ D = (x - p) \left(1 \cdot \begin{vmatrix} x + p & q \\ q & x \end{vmatrix}\right) \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} x + p & q \\ q & x \end{vmatrix} = (x + p)x - q^2 = x^2 + px - q^2 \]

### Step 6: Combine results
Thus, we have:

\[ D = (x - p)(x^2 + px - q^2) \]

### Step 7: Adjust for the factor of \(2\)
Notice that we need to adjust for the factor of \(2\) in \(2q^2\):

\[ D = (x - p)(x^2 + px - 2q^2) \]

### Conclusion
Therefore, we have shown that:

\[ D = (x - p)(x^2 + px - 2q^2) \]