Suppose a,b,c are distinct real number and `Delta=|(a,a^(2),b+c),(b,b^(2),c+a),(c,c^(2),a+b)|=0`
Then `a+b+c` equals

Options

A. `-1`
B. 2
C. 0
D. `-5`

Answer

To solve the problem, we need to evaluate the determinant given by:

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

and find the value of \( a + b + c \) given that \( \Delta = 0 \).

### Step 1: Expand the Determinant
We will use the formula for the determinant of a 3x3 matrix:

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

### Step 2: Calculate the 2x2 Determinants
1. Calculate \( \begin{vmatrix} b^2 & c+a \\ c^2 & a+b \end{vmatrix} \):
\[ = b^2(a+b) - (c+a)c^2 = ab^2 + b^2c - c^3 - ac^2 \]

2. Calculate \( \begin{vmatrix} b & c+a \\ c & a+b \end{vmatrix} \):
\[ = b(a+b) - c(c+a) = ab + b^2 - c^2 - ac \]

3. Calculate \( \begin{vmatrix} b & b^2 \\ c & c^2 \end{vmatrix} \):
\[ = b(c^2) - b^2(c) = bc^2 - b^2c = c(bc - b^2) \]

### Step 3: Substitute Back into the Determinant
Substituting these back into the determinant expression gives:

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

### Step 4: Set the Determinant to Zero
Since \( \Delta = 0 \), we can simplify the expression. However, we notice that the determinant being zero implies that the rows are linearly dependent.

### Step 5: Analyze Linear Dependence
The rows of the determinant can be expressed in terms of \( a + b + c \). If we assume \( a + b + c = 0 \), we can express one variable in terms of the others, say \( c = -a - b \).

### Step 6: Conclusion
Thus, if \( a + b + c = 0 \), the determinant will indeed equal zero, confirming that the condition \( \Delta = 0 \) holds true.

Therefore, the value of \( a + b + c \) is:

\[ \boxed{0} \]