Question

What is the value of xyz?

Statement 1: \( x^{a} = y^{b} = z^{c} \) and \( ab + bc + ca = 0 \) where a, b and c are non-zero integers
Statement 2: \( a^{x} = b, \, b^{y} = c, \, c^{z} = a \) where a, b and c are non-zero integers

• A. If the data in statement (1) alone is sufficient to answer the question
• B. If the data in statement (2) alone is sufficient to answer the question
• C. If the data in both the statements together are needed to answer the question
• D. If either statement (1) alone or statement (2) alone is sufficient to answer the question

Hint:

Try writing all three quantities in terms of one common base or one common exponent.

Answer 1

The Correct Option is D

Step 1: Understand what a single value of xyz would mean.
We need to check whether each statement pins xyz to one fixed number.

Step 2: Test statement 1 alone.
Let \( x^{a} = y^{b} = z^{c} = k \) for some nonzero k.
Then \( x = k^{1/a} \), \( y = k^{1/b} \), \( z = k^{1/c} \).
Multiplying gives \( xyz = k^{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} = k^{\frac{ab+bc+ca}{abc}} \).
Since \( ab + bc + ca = 0 \), the exponent is 0, so \( xyz = k^{0} = 1 \).
This holds for every valid k, so statement 1 alone fixes xyz = 1.

Step 3: Test statement 2 alone.
From \( a^{x} = b \), raise both sides to the power y: \( a^{xy} = b^{y} = c \).
Raise this new equation to the power z: \( a^{xyz} = c^{z} = a \).
So \( a^{xyz} = a^{1} \), and since a is a fixed nonzero base, the exponents must match: \( xyz = 1 \).
Statement 2 alone also fixes xyz = 1.

Final Answer:
Each statement alone is enough on its own to show xyz = 1. \[ \boxed{(d)} \]