If `a^(x)=b^(y)=c^(z)` and abc=1, then find `((x^(2))/(x^(2)-zy)+(y^(2))/(y^(2)-zx)+(z^(2))/(z^(2)-xy))`

Options

A. 3
B. 1
C. 0
D. x+y+z

Answer

To solve the problem, we start with the given equations and relationships.

### Step 1: Understand the given equations
We have:
1. \( a^x = b^y = c^z = k \) (where \( k \) is a constant)
2. \( abc = 1 \)

### Step 2: Express \( a, b, c \) in terms of \( k \)
From \( a^x = k \), we can express \( a \) as:
\[ a = k^{\frac{1}{x}} \]
Similarly, we can express \( b \) and \( c \):
\[ b = k^{\frac{1}{y}}, \quad c = k^{\frac{1}{z}} \]

### Step 3: Substitute \( a, b, c \) into the product equation
We know that:
\[ abc = k^{\frac{1}{x}} \cdot k^{\frac{1}{y}} \cdot k^{\frac{1}{z}} = k^{\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right)} = 1 \]
This implies:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \]

### Step 4: Rearranging the equation
From the equation \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \), we can multiply through by \( xyz \) to obtain:
\[ yz + xz + xy = 0 \]

### Step 5: Substitute into the expression we need to evaluate
We need to find:
\[ \frac{x^2}{x^2 - zy} + \frac{y^2}{y^2 - zx} + \frac{z^2}{z^2 - xy} \]

### Step 6: Substitute \( -zy, -zx, -xy \) using the relation \( xy + xz + yz = 0 \)
From the relation we derived, we can express:
- \( -zy = xy + xz \)
- \( -zx = xy + yz \)
- \( -xy = xz + yz \)

### Step 7: Substitute these values into the expression
Now substituting these into the expression:
\[ \frac{x^2}{x^2 + xy + xz} + \frac{y^2}{y^2 + xy + yz} + \frac{z^2}{z^2 + xz + yz} \]

### Step 8: Factor the denominators
We can factor the denominators:
\[ = \frac{x^2}{x(x + y + z)} + \frac{y^2}{y(y + x + z)} + \frac{z^2}{z(z + x + y)} \]

### Step 9: Simplify the expression
This simplifies to:
\[ = \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z} \]
Combining these fractions gives:
\[ = \frac{x + y + z}{x + y + z} = 1 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{1} \]