The value for `[axx(b+c),bxx(c-2a),cxx(a+3b)]` is equal to

Options

A. `[abc]^(2)`
B. `7[abc]^(2)`
C. `-5[axxb" "bxxc" "cxxa]`
D. none of these

Answer

To solve the problem, we need to evaluate the expression \([a \times (b+c), b \times (c-2a), c \times (a+3b)]\). This involves calculating the scalar triple product of three vectors derived from the cross products of the given vectors.

### Step-by-Step Solution:

1. **Expand the Cross Products**:
We start by expanding each of the terms in the expression:
\[ a \times (b+c) = a \times b + a \times c \]
\[ b \times (c-2a) = b \times c - 2(b \times a) \]
\[ c \times (a+3b) = c \times a + 3(c \times b) \]

2. **Combine the Results**:
Now we can combine these results:
\[ [a \times b + a \times c, b \times c - 2(b \times a), c \times a + 3(c \times b)] \]

3. **Let’s Define Variables**:
Let:
\[ x = a \times b, \quad y = b \times c, \quad z = c \times a \]
Then we can rewrite our expression in terms of \(x\), \(y\), and \(z\):
\[ [x + a \times c, y - 2x, z + 3y] \]

4. **Evaluate the Scalar Triple Product**:
The scalar triple product can be expressed as:
\[ x \cdot (y \times z) \]
We need to evaluate:
\[ (x + a \times c) \cdot ((y - 2x) \times (z + 3y)) \]

5. **Use Properties of the Cross Product**:
We know that if any two vectors in the scalar triple product are the same, the result is zero. Thus, we can simplify:
\[ x \cdot (y \times z) - 2x \cdot (y \times y) + 3y \cdot (x \times y) \]
Since \(y \times y = 0\), this term vanishes.

6. **Final Calculation**:
The remaining terms yield:
\[ x \cdot (y \times z) + 6(y \cdot (z \times x)) \]
By the properties of the scalar triple product, we can express this as:
\[ 7 \cdot (x \cdot (y \times z)) \]
This leads us to conclude that:
\[ [a \times (b+c), b \times (c-2a), c \times (a+3b)] = 7 \cdot (a \cdot (b \times c))^2 \]

### Conclusion:
Thus, the value of \([a \times (b+c), b \times (c-2a), c \times (a+3b)]\) is equal to \(7\) times the scalar triple product of \(a\), \(b\), and \(c\) squared.