Simplify the following expression. [(x − y)³ + (y − z)³ + (z − x)³] ⁄ [(x − y)(y − z)(z − x)]

Correct — A. Name the three brackets, then check what they add up to.

Let a = x − y, b = y − z, c = z − x
a + b + c = x − y + y − z + z − x = 0

When a + b + c = 0 the identity gives a³ + b³ + c³ = 3abc.

Numerator = 3(x − y)(y − z)(z − x)
Denominator = (x − y)(y − z)(z − x)

The product cancels, leaving 3 → option (a).

Why the others are wrong

• (b)
  2 — 2 would need a³ + b³ + c³ = 2abc, which is not the identity. The multiplier it supplies is 3, and it does not depend on x, y or z.
• (c)
  1 — 1 treats numerator and denominator as the same quantity. They differ by exactly the factor 3 — the cubes sum to three times the product, not to the product itself.
• (d)
  0 — 0 confuses a + b + c = 0 with a³ + b³ + c³ = 0. The cubes do not vanish; they collapse to 3abc, which is non-zero whenever x, y and z are distinct.

The identity behind this is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

The right-hand side dies the moment a + b + c = 0, which leaves a³ + b³ + c³ = 3abc.

Differences of the form (x − y), (y − z), (z − x) are the usual disguise for that condition: whatever x, y and z are, those three always sum to zero.

So the expression is 3 for every admissible x, y and z — it is a constant, not a value that happens to come out of one substitution.

Key facts

• a³ + b³ + c³ − 3abc factorises as (a + b + c)(a² + b² + c² − ab − bc − ca).
• If a + b + c = 0 then a³ + b³ + c³ = 3abc.
• (x − y) + (y − z) + (z − x) = 0 for all x, y and z, so the condition is automatic here.
• The given expression therefore equals 3 and is independent of x, y and z.

Common traps

• Reading a + b + c = 0 as forcing the sum of the cubes to be zero
• Expanding the three cubes term by term and losing a sign in (z − x)³

No directly related past PYQ was found.