`(vecbxxvecc)xx(veccxxveca)=`

Options

A. `[(veca,vecb,vecc)]vecc`
B. `[(veca,vecb,vecc)]vecb`
C. `[(veca,vecb,vecc)]veca`
D. `axx(vecbxxvecc)`

Explanation

To solve the expression \((\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a})\), we will use the vector triple product identity. The vector triple product identity states that for any vectors \(\mathbf{x}\), \(\mathbf{y}\), and \(\mathbf{z}\):

\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]

### Step 1: Identify the vectors
Let:
- \(\mathbf{x} = \mathbf{b}\)
- \(\mathbf{y} = \mathbf{c}\)
- \(\mathbf{z} = \mathbf{a}\)

### Step 2: Apply the vector triple product identity
Using the identity, we can rewrite the expression:

\[ (\mathbf{b} \times (\mathbf{c} \times \mathbf{a})) = (\mathbf{b} \cdot \mathbf{a}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} \]

### Step 3: Substitute back into the expression
Now, we need to find \((\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a})\):

Using the identity again, we can rewrite \((\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a})\):

\[ (\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a}) = (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \mathbf{c} - (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{c} \mathbf{a} \]

### Step 4: Evaluate the dot products
1. The term \((\mathbf{b} \times \mathbf{c}) \cdot \mathbf{c}\) is zero because \(\mathbf{b} \times \mathbf{c}\) is perpendicular to \(\mathbf{c}\).
2. Therefore, we are left with:

\[ (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \mathbf{c} \]

### Step 5: Write the final expression
The final expression can be simplified to:

\[ (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \mathbf{c} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \mathbf{c} \]

### Conclusion
Thus, the value of \((\mathbf{b} \times \mathbf{c}) \times (\mathbf{c} \times \mathbf{a})\) is:

\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \mathbf{c} \]