Suppose `a,b,c epsilon R` and let
`f(x)=|(0,a-x,b-x),(-a-x,0,c-x),(-b-x,-c-x,0)|` Then coefficient of `x^(2)` in f(x) is

Options

A. `-(a+b+c)`
B. `a+b+c`
C. `0`
D. `ab+bc+ca`

Answer

To find the coefficient of \( x^2 \) in the function

\[ f(x) = \left| \begin{array}{ccc} 0 & a - x & b - x \\ -a - x & 0 & c - x \\ -b - x & -c - x & 0 \end{array} \right| \]

we will calculate the determinant step by step.

### Step 1: Write the determinant
We start with the determinant of the given matrix:

\[ f(x) = \left| \begin{array}{ccc} 0 & a - x & b - x \\ -a - x & 0 & c - x \\ -b - x & -c - x & 0 \end{array} \right| \]

### Step 2: Expand the determinant
We can use the cofactor expansion along the first row. The first row has a zero, so we only need to consider the second and third elements:

\[ f(x) = (a - x) \left| \begin{array}{cc} -a - x & c - x \\ -b - x & 0 \end{array} \right| - (b - x) \left| \begin{array}{cc} -a - x & 0 \\ -b - x & -c - x \end{array} \right| \]

### Step 3: Calculate the 2x2 determinants
Now we compute the two 2x2 determinants:

1. For the first determinant:

\[ \left| \begin{array}{cc} -a - x & c - x \\ -b - x & 0 \end{array} \right| = 0 \cdot (-a - x) - (c - x)(-b - x) = (c - x)(b + x) = cb + cx - bx - x^2 \]

2. For the second determinant:

\[ \left| \begin{array}{cc} -a - x & 0 \\ -b - x & -c - x \end{array} \right| = (-a - x)(-c - x) - 0 = (a + x)(c + x) = ac + ax + cx + x^2 \]

### Step 4: Substitute back into \( f(x) \)
Now substituting back into \( f(x) \):

\[ f(x) = (a - x)(cb + cx - bx - x^2) - (b - x)(ac + ax + cx + x^2) \]

### Step 5: Expand and combine like terms
Expanding both terms:

1. For the first term:

\[ (a - x)(cb + cx - bx - x^2) = acb + acx - abx - ax^2 - x(cb + cx - bx - x^2) \]

2. For the second term:

\[ -(b - x)(ac + ax + cx + x^2) = -bac - b^2x - bcx - bx^2 + acx + ax^2 + cx^2 + x^3 \]

### Step 6: Collect coefficients of \( x^2 \)
Now we need to collect the coefficients of \( x^2 \):

From the first term, the coefficient of \( x^2 \) is \( -1 \) (from \( -x^2 \)).

From the second term, the coefficient of \( x^2 \) is \( -b + a + c \).

Combining these gives:

\[ \text{Coefficient of } x^2 = -1 - b + a + c \]

### Step 7: Final result
Setting the coefficients equal to zero to find the coefficient of \( x^2 \):

\[ \text{Coefficient of } x^2 = 0 \]

Thus, the coefficient of \( x^2 \) in \( f(x) \) is:

\[ \boxed{0} \]