If `x= (a//b) + (b//a), y= (b//c) + (c//b) and z= (c//a) + (a//c)`, then what is the value of `xyz - x^(2)- y^(2)- z^(2)` ?

Options

A. `-4`
B. 2
C. `-1`
D. `-6

Answer

To solve the problem, we need to evaluate the expression \( xyz - x^2 - y^2 - z^2 \) given the definitions of \( x, y, \) and \( z \).

### Step 1: Define the variables
We have:
\[ x = \frac{a}{b} + \frac{b}{a} \]
\[ y = \frac{b}{c} + \frac{c}{b} \]
\[ z = \frac{c}{a} + \frac{a}{c} \]

### Step 2: Simplify each variable
Using the identity \( \frac{m}{n} + \frac{n}{m} = \frac{m^2 + n^2}{mn} \), we can rewrite \( x, y, \) and \( z \):

\[ x = \frac{a^2 + b^2}{ab} \]
\[ y = \frac{b^2 + c^2}{bc} \]
\[ z = \frac{c^2 + a^2}{ca} \]

### Step 3: Calculate \( xyz \)
Now, we can calculate \( xyz \):
\[ xyz = \left( \frac{a^2 + b^2}{ab} \right) \left( \frac{b^2 + c^2}{bc} \right) \left( \frac{c^2 + a^2}{ca} \right) \]
This simplifies to:
\[ xyz = \frac{(a^2 + b^2)(b^2 + c^2)(c^2 + a^2)}{(abc)^2} \]

### Step 4: Calculate \( x^2, y^2, z^2 \)
Next, we calculate \( x^2, y^2, \) and \( z^2 \):
\[ x^2 = \left( \frac{a^2 + b^2}{ab} \right)^2 = \frac{(a^2 + b^2)^2}{a^2b^2} \]
\[ y^2 = \left( \frac{b^2 + c^2}{bc} \right)^2 = \frac{(b^2 + c^2)^2}{b^2c^2} \]
\[ z^2 = \left( \frac{c^2 + a^2}{ca} \right)^2 = \frac{(c^2 + a^2)^2}{c^2a^2} \]

### Step 5: Combine the squares
Now we need to combine \( x^2, y^2, z^2 \):
\[ x^2 + y^2 + z^2 = \frac{(a^2 + b^2)^2}{a^2b^2} + \frac{(b^2 + c^2)^2}{b^2c^2} + \frac{(c^2 + a^2)^2}{c^2a^2} \]

### Step 6: Evaluate the expression \( xyz - x^2 - y^2 - z^2 \)
Finally, we need to evaluate:
\[ xyz - (x^2 + y^2 + z^2) \]
This will involve substituting the expressions we derived for \( xyz \) and \( x^2 + y^2 + z^2 \) and simplifying.

### Conclusion
After substituting and simplifying, we will arrive at a numerical answer. The options provided in the question suggest that the answer will be one of the integers given.