For distinct real numbers x, y and z, evaluate the expression ((x − y)^3 + (y − z)^3 + (z − x)^3) ÷ (3(x − y)(y − z)(z − x)).

Introduction / Context:
This algebraic identity problem involves symmetric expressions in three variables x, y and z. The numerator is a sum of three cubes of pairwise differences, while the denominator is three times the product of those same differences. At first glance, it may look complicated, but it actually relates to a well known identity involving a^3 + b^3 + c^3 − 3abc. The question checks understanding of identities and ability to apply them in a nonstandard form.

Given Data / Assumptions:

• x, y and z are real numbers, all distinct so that denominators are nonzero.
• The numerator is (x − y)^3 + (y − z)^3 + (z − x)^3.
• The denominator is 3(x − y)(y − z)(z − x).
• We need to simplify the entire fraction to a single, simple value.

Concept / Approach:
The expression resembles the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)[(a − b)^2 + (b − c)^2 + (c − a)^2]/2. In this problem, we can set a = x − y, b = y − z and c = z − x. Notice that a + b + c equals zero in this case, which considerably simplifies the identity. Once we understand that the sum of cubes equals 3abc whenever a + b + c = 0, the fraction simplifies immediately.

Step-by-Step Solution:
Step 1: Let a = x − y, b = y − z and c = z − x.Step 2: Observe that a + b + c = (x − y) + (y − z) + (z − x) = 0.Step 3: Recall the identity a^3 + b^3 + c^3 − 3abc = (a + b + c)[(a − b)^2 + (b − c)^2 + (c − a)^2]/2.Step 4: When a + b + c = 0, the right side becomes zero, so a^3 + b^3 + c^3 − 3abc = 0.Step 5: Therefore a^3 + b^3 + c^3 = 3abc.Step 6: Substitute back a = x − y, b = y − z and c = z − x to get (x − y)^3 + (y − z)^3 + (z − x)^3 = 3(x − y)(y − z)(z − x).Step 7: Hence the given fraction becomes [3(x − y)(y − z)(z − x)] ÷ [3(x − y)(y − z)(z − x)].Step 8: The numerator and denominator are identical nonzero expressions, so the fraction simplifies to 1.

Verification / Alternative check:
To verify numerically, pick distinct values such as x = 1, y = 2 and z = 4. Compute the differences: x − y = −1, y − z = −2 and z − x = 3. Then the numerator is (−1)^3 + (−2)^3 + 3^3 = −1 − 8 + 27 = 18. The denominator is 3(−1)(−2)(3) = 3 × 6 = 18. The fraction is 18/18 = 1, confirming the result for this specific choice of x, y and z. Since the derivation used an identity, it holds for all distinct real numbers x, y and z.

Why Other Options Are Wrong:
The option 0 would require the numerator to vanish while the denominator stays nonzero, which contradicts the proven identity. The value 1/3 or 3 could only appear if the numerator were a constant multiple of the denominator, which it is not in this case after using the correct identity. The option −1 would require a sign mismatch between numerator and denominator, which does not occur. Only 1 is consistent with both algebraic reasoning and numerical verification.

Common Pitfalls:
Students sometimes attempt to expand each cube fully and then factor the resulting polynomial, which is tedious and error prone. Another frequent mistake is misremembering the a^3 + b^3 + c^3 − 3abc identity or failing to notice that a + b + c equals zero for these particular differences. Recognising this special condition immediately simplifies the problem and avoids long calculations.

Final Answer:
The expression ((x − y)^3 + (y − z)^3 + (z − x)^3) ÷ (3(x − y)(y − z)(z − x)) simplifies to 1.