For all positive numbers x,y,z the product `((1)/(x+y+z))((1)/(x)+(1)/(y)+(1)/(z))((1)/(xy+yz+zx))((1)/(xy)+(1)/(yz)+(1)/(zx))` equals

Options

A. `x^(-2)y^(-2)z^(-2)`
B. `x^(-2)+y^(-2)+z^(-2)`
C. `(x+y+z)^(-2)`
D. `(1)/(xyz)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \left(\frac{1}{x+y+z}\right) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \left(\frac{1}{xy+yz+zx}\right) \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \]

Let's break this down step by step.

### Step 1: Rewrite the second term
The second term can be rewritten using a common denominator. The common denominator for \(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\) is \(xyz\):

\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{yz + xz + xy}{xyz} \]

### Step 2: Rewrite the fourth term
Similarly, the fourth term can also be rewritten using a common denominator, which is again \(xyz\):

\[ \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = \frac{z + x + y}{xyz} \]

### Step 3: Substitute back into the expression
Now substituting these back into the original expression, we have:

\[ \left(\frac{1}{x+y+z}\right) \left(\frac{yz + xz + xy}{xyz}\right) \left(\frac{1}{xy + yz + zx}\right) \left(\frac{z + x + y}{xyz}\right) \]

### Step 4: Combine the fractions
Now we can combine the fractions:

\[ = \frac{(yz + xz + xy)(z + x + y)}{(x+y+z)(xy + yz + zx)(xyz^2)} \]

### Step 5: Simplify the expression
Notice that \(yz + xz + xy\) is the same as \(xy + xz + yz\), so we can simplify:

\[ = \frac{(yz + xz + xy)(z + x + y)}{(x+y+z)(xy + yz + zx)(xyz^2)} \]

### Step 6: Cancel terms
The numerator and denominator can be simplified further. The term \(x+y+z\) in the denominator will cancel with the corresponding term in the numerator. Thus, we have:

\[ = \frac{1}{xyz} \]

### Step 7: Final expression
Finally, we can express this as:

\[ = \frac{1}{x^2y^2z^2} \]

This can be rewritten using negative exponents:

\[ = x^{-2}y^{-2}z^{-2} \]

### Conclusion
Thus, the final answer is:

\[ \boxed{x^{-2}y^{-2}z^{-2}} \]