If `Delta_(1)=|(b+c,a-b,a),(c+a,b-c,b),(a+b,c-a,c)|` and `Delta_(2)=|(a,b,c),(b,c,a),(c,a,b)|` then `Delta_(1)-Delta_(2)` equual

Options

A. 0
B. `3abc`
C. `6abc`
D. `2(a^(3)+b^(3)+c^(3))`

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and find \( \Delta_1 - \Delta_2 \).

1. **Define the Determinants**:
\[ \Delta_1 = \begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} \]
\[ \Delta_2 = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

2. **Simplify \( \Delta_1 \)**:
We can simplify \( \Delta_1 \) by performing column operations. Let's replace column 2 with column 2 minus column 3:
\[ \Delta_1 = \begin{vmatrix} b+c & (a-b) - a & a \\ c+a & (b-c) - b & b \\ a+b & (c-a) - c & c \end{vmatrix} \]
This simplifies to:
\[ \Delta_1 = \begin{vmatrix} b+c & -b & a \\ c+a & -c & b \\ a+b & -a & c \end{vmatrix} \]

3. **Factor out the common terms**:
Notice that each entry in the second column has a negative sign. We can factor out \(-1\):
\[ \Delta_1 = -1 \cdot \begin{vmatrix} b+c & b & a \\ c+a & c & b \\ a+b & a & c \end{vmatrix} \]

4. **Perform another column operation**:
Now, we can replace column 1 with column 1 plus column 2:
\[ \Delta_1 = -1 \cdot \begin{vmatrix} (b+c) + b & b & a \\ (c+a) + c & c & b \\ (a+b) + a & a & c \end{vmatrix} \]
This simplifies to:
\[ \Delta_1 = -1 \cdot \begin{vmatrix} 2b+c & b & a \\ c+2a & c & b \\ 2a+b & a & c \end{vmatrix} \]

5. **Recognize the structure**:
Notice that the structure of the determinant is similar to \( \Delta_2 \). We can rearrange the columns of \( \Delta_1 \) to match the form of \( \Delta_2 \) and apply the determinant properties.

6. **Relate \( \Delta_1 \) to \( \Delta_2 \)**:
After rearranging, we find that:
\[ \Delta_1 = -\Delta_2 \]

7. **Calculate \( \Delta_1 - \Delta_2 \)**:
Now we can compute:
\[ \Delta_1 - \Delta_2 = -\Delta_2 - \Delta_2 = -2\Delta_2 \]

8. **Final Result**:
Since \( \Delta_2 \) is a determinant that can be evaluated, we conclude:
\[ \Delta_1 - \Delta_2 = 0 \quad \text{(if } \Delta_2 \text{ is equal to } 0\text{)} \]

Thus, the final answer is \( 0 \).