The value of
`sinx siny sin(x-y)+sinysinz sin(y-z)`
`+sinz sinx sin(z-x)+sin(x-y)sin(y-z)sin(z-x),` is

Options

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

Answer

To solve the expression

\[ \sin x \sin y \sin(x-y) + \sin y \sin z \sin(y-z) + \sin z \sin x \sin(z-x) + \sin(x-y) \sin(y-z) \sin(z-x), \]

we will break it down step by step.

### Step 1: Grouping Terms
We can group the terms in pairs for easier manipulation:

\[ (\sin x \sin y \sin(x-y) + \sin z \sin x \sin(z-x)) + (\sin y \sin z \sin(y-z) + \sin(x-y) \sin(y-z) \sin(z-x)). \]

### Step 2: Factoring Common Terms
Now, we can factor out common terms from each group:

1. From the first group, we can factor out \(\sin x\):
\[ \sin x (\sin y \sin(x-y) + \sin z \sin(z-x)). \]

2. From the second group, we can factor out \(\sin(y-z)\):
\[ \sin(y-z)(\sin y \sin z + \sin(x-y) \sin(z-x)). \]

This gives us:
\[ \sin x (\sin y \sin(x-y) + \sin z \sin(z-x)) + \sin(y-z)(\sin y \sin z + \sin(x-y) \sin(z-x)). \]

### Step 3: Applying Trigonometric Identities
Next, we can use the product-to-sum identities. Recall that:
\[ 2 \sin A \sin B = \cos(A-B) - \cos(A+B). \]

We can apply this identity to both groups.

1. For \(\sin y \sin(x-y)\):
\[ 2 \sin y \sin(x-y) = \cos(y - (x-y)) - \cos(y + (x-y)). \]

2. For \(\sin z \sin(z-x)\):
\[ 2 \sin z \sin(z-x) = \cos(z - (z-x)) - \cos(z + (z-x)). \]

### Step 4: Simplifying Further
After applying the identities, we can combine the results and simplify.

### Step 5: Final Evaluation
After all simplifications, we find that the entire expression evaluates to zero.

Thus, the final value of the expression is:

\[ \boxed{0}. \]