Question

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:

• A. A and B only
• B. A and C only
• C. B and C only
• D. A, B and C only

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze the provided equations and verify which options are true. 

The given equations are:

\(x - \frac{1}{x} = y\), \(y - \frac{1}{y} = z\), \(z - \frac{1}{z} = x\).

Now, let's consider the options one by one:

1. Option A: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
   • For \(y\), we have: \(\frac{1}{x} = x - y\)
   • For \(z\), we have: \(\frac{1}{y} = y - z\)
   • For \(x\), we have: \(\frac{1}{z} = z - x\)
2. Option B: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=8\)
   • \(\left(\frac{1}{x}\right)^2 = (x - y)^2\)
   • \(\left(\frac{1}{y}\right)^2 = (y - z)^2\)
   • \(\left(\frac{1}{z}\right)^2 = (z - x)^2\)
3. Option C: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=-3\)
   • \(\frac{1}{xy} = \frac{1}{x}\frac{1}{y} = (x-y)(y-z)\)
   • \(\frac{1}{yz} = \frac{1}{y}\frac{1}{z} = (y-z)(z-x)\)
   • \(\frac{1}{zx} = \frac{1}{z}\frac{1}{x} = (z-x)(x-y)\)

Conclusion: After evaluating each option, both options A and C are correct, thus the correct answer is "A and C only".