If \(x - \frac{1}{x} = y\), \(y - \frac{1}{y} = z\) and \(z - \frac{1}{z} = x\), then which of the…

A and C only. Adding the three equations gives $x$ $+ y$ $+ z$ $- (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ $= x + y + z$, so the reciprocals sum to 0. Squaring each equation gives $x^{2} - 2 + \frac{1}{x^{2}}$ $= y^{2}$ and so on; adding these, the squares cancel and $\sum \frac{1}{x^{2}} = 6$, not 8. Then $0$ $= (\sum \frac{1}{x})^{2}$ $= 6 + 2\sum \frac{1}{xy}$, so $\sum \frac{1}{xy} = -3$.