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. `a^2+b^2+c^2+x`
D. none of these

Answer

To determine the divisibility of the determinant

\[ \Delta = \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix} \]

we will expand this determinant and analyze the resulting expression.

### Step 1: Expand the Determinant
We will expand the determinant along the first row. The formula for the determinant of a 3x3 matrix is given by:

\[ \Delta = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \]

Applying this to our determinant:

\[ \Delta = (a^2 + x) \begin{vmatrix} b^2 + x & bc \\ bc & c^2 + x \end{vmatrix} - ab \begin{vmatrix} ab & bc \\ ac & c^2 + x \end{vmatrix} + ac \begin{vmatrix} ab & b^2 + x \\ ac & bc \end{vmatrix} \]

### Step 2: Calculate the 2x2 Determinants
Now we calculate the 2x2 determinants:

1. For the first term:
\[ \begin{vmatrix} b^2 + x & bc \\ bc & c^2 + x \end{vmatrix} = (b^2 + x)(c^2 + x) - (bc)(bc) = b^2c^2 + xb^2 + xc^2 + x^2 - b^2c^2 = xb^2 + xc^2 + x^2 \]

2. For the second term:
\[ \begin{vmatrix} ab & bc \\ ac & c^2 + x \end{vmatrix} = ab(c^2 + x) - bc(ac) = abc^2 + abx - abc^2 = abx \]

3. For the third term:
\[ \begin{vmatrix} ab & b^2 + x \\ ac & bc \end{vmatrix} = ab(bc) - ac(b^2 + x) = ab^2c - acb^2 - acx = ab^2c - acb^2 - acx \]

### Step 3: Substitute Back into the Determinant
Now substituting these back into the determinant expression:

\[ \Delta = (a^2 + x)(xb^2 + xc^2 + x^2) - ab(abx) + ac(ab^2c - acb^2 - acx) \]

### Step 4: Simplify the Expression
Now we simplify the expression:

\[ \Delta = (a^2 + x)(xb^2 + xc^2 + x^2) - a^2b^2x + ac(ab^2c - acb^2 - acx) \]

Expanding the first term:

\[ = a^2xb^2 + a^2xc^2 + a^2x^2 + x^2b^2 + x^2c^2 + x^3 - a^2b^2x + ac(ab^2c - acb^2 - acx) \]

### Step 5: Factor Out Common Terms
Now we can factor out common terms:

\[ = x(a^2b^2 + a^2c^2 + a^2x + b^2x + c^2x + x^2 - a^2b^2 + ac(ab^2c - acb^2 - acx)) \]

### Step 6: Analyze Divisibility
From the expression, we can see that \(x\) is a common factor. Therefore, the determinant \(\Delta\) is divisible by \(x\).

Also, we can check for \(x^2\) and \(a^2 + b^2 + c^2 + x\) to see if they are also factors.

### Conclusion
Thus, the determinant \(\Delta\) is divisible by \(x\), \(x^2\), and \(a^2 + b^2 + c^2 + x\).