the determinant `Delta=|[a^2+x, ab, ac] , [ab, b^2+x, bc] , [ac, bc, c^2+x]|` is divisible by

Options

A. x
B. `x^(2)`
C. `x^(3)`
D. none of these

Explanation

To solve the determinant \( \Delta = \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix} \) and determine its divisibility by \( x \), \( x^2 \), or \( x^3 \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix} \]

### Step 2: Multiply the First Column by \( a \)
We multiply the first column by \( a \):
\[ \Delta = a \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix} \]

### Step 3: Apply Column Transformation
Next, we perform a column transformation by changing \( C_1 \) to \( C_1 + bC_2 + cC_3 \):
\[ \Delta = a \begin{vmatrix} a^3 + ax + ab^2 + ac^2 & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix} \]

### Step 4: Factor Out Common Terms
Now, we can factor out common terms from the first column:
\[ \Delta = a \cdot (a^2 + b^2 + c^2 + x) \begin{vmatrix} 1 & b & c \\ b & 0 & 0 \\ c & 0 & 0 \end{vmatrix} \]

### Step 5: Calculate the Remaining Determinant
Now we calculate the remaining determinant:
\[ \begin{vmatrix} 1 & b & c \\ b & 0 & 0 \\ c & 0 & 0 \end{vmatrix} = 1 \cdot (0 \cdot 0 - 0 \cdot 0) - b(0 \cdot 0 - c \cdot 0) + c(b \cdot 0 - 0 \cdot 0) = 0 \]
Thus, the determinant simplifies to:
\[ \Delta = (a^2 + b^2 + c^2 + x) \cdot 0 = 0 \]

### Step 6: Conclusion on Divisibility
Since \( \Delta = 0 \), it is divisible by any power of \( x \), including \( x \), \( x^2 \), and \( x^3 \).

### Final Answer
The determinant \( \Delta \) is divisible by \( x \), \( x^2 \), and \( x^3 \).