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

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then find the ratio \( \frac{\Delta_1}{\Delta_2} \).

### Step 1: Evaluate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]

We can use the property of determinants that allows us to split columns. We can rewrite the first column as:
\[ \Delta_1 = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]
This can be rewritten by splitting the first column into two parts:
\[ = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \]

### Step 2: Simplify \( \Delta_1 \)

Next, we perform column operations. We can replace the third column with the third column minus the first column:
\[ \Delta_1 = \begin{vmatrix} b+c & c+a & (a+b) - (b+c) \\ c+a & a+b & (b+c) - (c+a) \\ a+b & b+c & (c+a) - (a+b) \end{vmatrix} \]
This simplifies to:
\[ = \begin{vmatrix} b+c & c+a & a-c \\ c+a & a+b & b-a \\ a+b & b+c & c-b \end{vmatrix} \]

### Step 3: Further Simplification

Next, we can perform additional column operations to simplify further. We can replace the second column with the second column minus the first column:
\[ = \begin{vmatrix} b+c & (c+a)-(b+c) & a-c \\ c+a & (a+b)-(c+a) & b-a \\ a+b & (b+c)-(a+b) & c-b \end{vmatrix} \]
This gives us:
\[ = \begin{vmatrix} b+c & a-b & a-c \\ c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix} \]

### Step 4: Recognizing the Structure

Now we can see that the determinant has a structure similar to \( \Delta_2 \):
\[ \Delta_2 = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 5: Relating \( \Delta_1 \) and \( \Delta_2 \)

We can show that \( \Delta_1 \) can be expressed in terms of \( \Delta_2 \) after performing a series of row and column swaps. Each swap changes the sign of the determinant, and since we perform an even number of swaps, we find that:
\[ \Delta_1 = 2 \Delta_2 \]

### Step 6: Calculate the Ratio

Thus, we can find the ratio:
\[ \frac{\Delta_1}{\Delta_2} = \frac{2 \Delta_2}{\Delta_2} = 2 \]

### Final Answer

\[ \frac{\Delta_1}{\Delta_2} = 2 \]