SSC CGL 2025 Tier-I · 17 September 2025, Shift 3 · Quantitative Aptitude · Question 6 of 100

We transcribed this text from the pictures on a candidate's response sheet for this shift; the answer is the key SSC colour-marked on that sheet. Solution last reviewed 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

Given x + y + z = 0, and x³ + y³ + z³ = 3xyz, evaluate (x − y)³ + (y − z)³ + (z − x)³.

(a)0
(b)9(x − y) (y − z) (z − x)
(c)27xyz
(d)3(x − y)(y − z)(z − x)

Answer

Why

Correct — D. Treat the three brackets as three numbers and add them.

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

When a + b + c = 0: a³ + b³ + c³ = 3abc
So the expression = 3(x − y)(y − z)(z − x) → option (d)

Why the others are wrong

(a)
0 — 0 is the sum of the brackets, not of their cubes. For x = 1, y = 2, z = −3 the expression is (−1)³ + 5³ + (−4)³ = 60.
(b)
9(x − y) (y − z) (z − x) — 9(x − y)(y − z)(z − x) has three times the right coefficient. For x = 1, y = 2, z = −3 it gives 180, while the expression is 60.
(c)
27xyz — 27xyz does not match the expression. For x = 1, y = 2, z = −3 it is 27 × (−6) = −162, while the expression is 60.

Concept

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Whenever a + b + c = 0 the right side vanishes, so a³ + b³ + c³ = 3abc.

The differences (x − y), (y − z) and (z − x) always add to zero, so the sum of their cubes is always 3(x − y)(y − z)(z − x).

The stem's two conditions, x + y + z = 0 and x³ + y³ + z³ = 3xyz, are never used. The result holds for any x, y, z, because the brackets cancel on their own.

The second condition also follows from the first, by the same identity applied to x, y and z.

Key facts

a³ + b³ + c³ − 3abc = (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 every x, y, z.

Study next

Evaluating sums such as 19³ + 20³ − 39³ with the a + b + c = 0 identitya² + b² + c² − ab − bc − ca = ½[(a − b)² + (b − c)² + (c − a)²]Factorising a³ + b³ + c³ − 3abc when a + b + c ≠ 0

Common traps

Expanding all three cubes term by term. It works, but it is slow and invites sign slips.
Assuming the answer must use the given x + y + z = 0. The brackets sum to zero for any x, y, z.

The same identity settles 17 Sep 2025, 16:00, Quant Q.21: 19 + 20 − 39 = 0, so 19³ + 20³ − 39³ = 3 × 19 × 20 × (−39) = −44460, and adding 118 gives −44342.

In 19 Sep 2025, 09:00, Quant Q.25, 0.5 + 0.1 − 0.6 = 0, so the numerator equals 3 × 0.5 × 0.1 × (−0.6) and the quotient is −1.

Related PYQs

No directly related past PYQ was found.

Topic: Basic algebraic identities of School Algebra & Elementary surds — all 147 SSC CGL questions on it, every solved shift.