If `a^(x) =b^(y) =c^(z)` and abc =1 then `xy+yz+zx` is equal to which of the following?

Options

A. xyz
B. `x+y+z`
C. 0
D. 1

Answer

To solve the problem, we start with the given equations and conditions:

1. **Given**: \( a^x = b^y = c^z = k \) and \( abc = 1 \).

2. **Expressing a, b, and c in terms of k**:
- From \( a^x = k \), we can express \( a \) as:
\[ a = k^{\frac{1}{x}} \]
- From \( b^y = k \), we can express \( b \) as:
\[ b = k^{\frac{1}{y}} \]
- From \( c^z = k \), we can express \( c \) as:
\[ c = k^{\frac{1}{z}} \]

3. **Substituting into the condition \( abc = 1 \)**:
\[ abc = k^{\frac{1}{x}} \cdot k^{\frac{1}{y}} \cdot k^{\frac{1}{z}} = k^{\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)} = 1 \]
Since \( k^0 = 1 \), we have:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \]

4. **Finding \( xy + yz + zx \)**:
- We know that:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \]
- This can be rewritten as:
\[ \frac{yz + zx + xy}{xyz} = 0 \]
- Therefore, the numerator must be zero:
\[ yz + zx + xy = 0 \]

5. **Conclusion**:
- Thus, we find that:
\[ xy + yz + zx = 0 \]

### Final Answer:
The value of \( xy + yz + zx \) is **0**.