If `x=sin(alpha-beta)*sin(gamma-delta), y=sin(beta-gamma)*sin(alpha-delta), z=sin(gamma-alpha)*sin(beta-delta)`, then :

Options

A. `x+y+z=0`
B. `x^(3)+y^(3)+z^(3)=3xyz`
C. `x+y-z=0`
D. `x^(3)+y^(3)-z^(3)=3xyz`

Answer

To solve the problem step by step, we start with the given expressions for \( x \), \( y \), and \( z \):

1. **Given Expressions**:
\[ x = \sin(\alpha - \beta) \cdot \sin(\gamma - \delta) \]
\[ y = \sin(\beta - \gamma) \cdot \sin(\alpha - \delta) \]
\[ z = \sin(\gamma - \alpha) \cdot \sin(\beta - \delta) \]

2. **Using the Product-to-Sum Formula**:
We apply the product-to-sum identities for sine:
\[ \sin A \cdot \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]

3. **Calculating \( 2x \)**:
\[ 2x = 2 \sin(\alpha - \beta) \cdot \sin(\gamma - \delta) = \cos(\alpha - \beta - (\gamma - \delta)) - \cos(\alpha - \beta + (\gamma - \delta)) \]
Simplifying:
\[ 2x = \cos(\alpha - \beta - \gamma + \delta) - \cos(\alpha - \beta + \gamma - \delta) \]

4. **Calculating \( 2y \)**:
\[ 2y = 2 \sin(\beta - \gamma) \cdot \sin(\alpha - \delta) = \cos(\beta - \gamma - (\alpha - \delta)) - \cos(\beta - \gamma + (\alpha - \delta)) \]
Simplifying:
\[ 2y = \cos(\beta - \gamma - \alpha + \delta) - \cos(\beta - \gamma + \alpha - \delta) \]

5. **Calculating \( 2z \)**:
\[ 2z = 2 \sin(\gamma - \alpha) \cdot \sin(\beta - \delta) = \cos(\gamma - \alpha - (\beta - \delta)) - \cos(\gamma - \alpha + (\beta - \delta)) \]
Simplifying:
\[ 2z = \cos(\gamma - \alpha - \beta + \delta) - \cos(\gamma - \alpha + \beta - \delta) \]

6. **Adding \( 2x + 2y + 2z \)**:
Now we add these three equations:
\[ 2x + 2y + 2z = [\cos(\alpha - \beta - \gamma + \delta) - \cos(\alpha - \beta + \gamma - \delta)] + [\cos(\beta - \gamma - \alpha + \delta) - \cos(\beta - \gamma + \alpha - \delta)] + [\cos(\gamma - \alpha - \beta + \delta) - \cos(\gamma - \alpha + \beta - \delta)] \]

7. **Using the Property of Cosine**:
We can use the property that \( \cos(-\theta) = \cos(\theta) \) to simplify the terms. When we combine these, we find that they cancel out in pairs, leading to:
\[ 2x + 2y + 2z = 0 \]

8. **Final Result**:
From the equation \( 2x + 2y + 2z = 0 \), we can conclude that:
\[ x + y + z = 0 \]
Using the identity \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \), since \( x + y + z = 0 \), we have:
\[ x^3 + y^3 + z^3 - 3xyz = 0 \]
Therefore, we conclude:
\[ x^3 + y^3 + z^3 = 3xyz \]