If $$x - \frac{1}{x} = y, y - \frac{1}{y} = z, z - \frac{1}{z} = x$$, there which of the following equations are true?
A. $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$$
B. $$\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}} = 8$$
C. $$\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = -3$$
Choose the correct answer from the options given below:

It is given,

$$x-\frac{1}{x}=y$$, $$y-\frac{1}{y}=z$$ and $$z-\frac{1}{z}=x$$

Adding all the given equations, we get

$$x\ +\ y\ +\ z\ -\ \frac{1}{x}-\frac{1}{y}-\frac{1}{z}=y+z+x$$

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

Therefore, equation A is correct.

$$x-\frac{1}{x}=y$$, $$y-\frac{1}{y}=z$$ and $$z-\frac{1}{z}=x$$

Squaring on both the sides and adding three resultant equations, we get

$$x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2+z^2+\frac{1}{z^2}-2=y^2+z^2+x^2$$

$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6$$

Therefore, equation B is incorrect.

$$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2=0$$

On expanding, $$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz})=0$$

We know that $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=6$$

So, $$2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) = -6$$

$$(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}) = -3$$

Therefore, equation C is correct.

The answer is option B.