Let
`Delta=|(a,x,x),(x,b,x),(x,x,c)|`
and `f(x)=(x-a)(x-b)(x-c)`
Determinant `Delta` is equal to

Options

A. `f(x)-x^(3)`
B. `f'(x)`
C. `xf'(x)-f(x)`
D. `f'(x)-xf''(x)`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} a & x & x \\ x & b & x \\ x & x & c \end{vmatrix} \), we will follow a step-by-step approach.

### Step 1: Write the Determinant
We start with the determinant as given:

\[ \Delta = \begin{vmatrix} a & x & x \\ x & b & x \\ x & x & c \end{vmatrix} \]

### Step 2: Apply Row Operations
To simplify the determinant, we can perform row operations. Let's subtract the second column from the first column:

\[ \Delta = \begin{vmatrix} a - x & x & x \\ x - x & b & x \\ x - x & x & c \end{vmatrix} = \begin{vmatrix} a - x & x & x \\ 0 & b & x \\ 0 & x & c \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant along the first column:

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

Calculating the 2x2 determinant:

\[ \begin{vmatrix} b & x \\ x & c \end{vmatrix} = bc - x^2 \]

Thus, we have:

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

### Step 4: Expand the Expression
Now we expand the expression:

\[ \Delta = (a - x)(bc - x^2) = abc - ax^2 - bcx + x^3 \]

### Step 5: Rearranging the Terms
Rearranging the terms gives us:

\[ \Delta = x^3 - (a + b + c)x^2 + ab + ac + bc \]

### Final Expression
Thus, the final expression for the determinant \( \Delta \) is:

\[ \Delta = x^3 - (a + b + c)x^2 + (ab + ac + bc) \]