If `x =a- b, y = b- c, z =c- a,` then the numeric value of the algebric expression `x ^(3) + y ^(3) + z ^(3) - 3 xyz` will be :

Options

A. `a + b + c`
B. 0
C. `4 (a + b+ c)`
D. 3 abc

Answer

To solve the problem, we need to evaluate the expression \( x^3 + y^3 + z^3 - 3xyz \) given the definitions of \( x \), \( y \), and \( z \):

1. **Given Definitions**:
- \( x = a - b \)
- \( y = b - c \)
- \( z = c - a \)

2. **Calculate \( x + y + z \)**:
\[ x + y + z = (a - b) + (b - c) + (c - a) \]
Simplifying this:
\[ = a - b + b - c + c - a \]
Notice that \( a \) cancels with \( -a \), \( b \) cancels with \( -b \), and \( c \) cancels with \( -c \):
\[ = 0 \]

3. **Substituting into the Expression**:
We use the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
Since we found that \( x + y + z = 0 \), we can substitute this into the identity:
\[ x^3 + y^3 + z^3 - 3xyz = 0 \cdot (x^2 + y^2 + z^2 - xy - yz - zx) \]
This simplifies to:
\[ = 0 \]

4. **Conclusion**:
Thus, the numeric value of the algebraic expression \( x^3 + y^3 + z^3 - 3xyz \) is \( 0 \).