SSC CGL 2024 Tier-I · 12 September 2024, Shift 3 · Quantitative Aptitude · Question 12 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

If a² + b² + c² = 2(a + c − 1), then the value of a³ + b³ + c³ =?

(a)0
(b)2
(c)4
(d)1

Answer

Why

Correct — B. Bring everything to one side and complete the squares.

a² + b² + c² = 2a + 2c − 2
a² − 2a + b² + c² − 2c + 2 = 0
(a² − 2a + 1) + b² + (c² − 2c + 1) = 0
(a − 1)² + b² + (c − 1)² = 0

Real squares add to zero only when each one is zero, so a = 1, b = 0, c = 1.

a³ + b³ + c³ = 1 + 0 + 1 = 2 → option (b)

Why the others are wrong

(a)
0 — Zero needs a = b = c = 0, and that triple breaks the given equation: the left side is 0 while 2(a + c − 1) is −2.
(c)
4 — Completing the square pins a and c at exactly 1, so there is nothing larger to cube. 4 is the value of 2(a + c), not of a³ + b³ + c³.
(d)
1 — 1 is what the triple (1, 0, 0) would give, but it fails the equation: 1 + 0 + 0 = 1 while 2(1 + 0 − 1) = 0. Both a and c are forced to 1, not just one of them.

Concept

The equation is a sum of squares equals zero in disguise. Over the real numbers a square is never negative, so if several squares add to zero every one of them must be zero — that is what turns one equation into three.

Getting there is mechanical: move the right-hand side across, then complete the square on each variable that carries a linear term. Here a and c each need a +1, and the constant supplies exactly those two.

b has no linear term, so its square stands alone and forces b = 0.

The stem is stored as an image in the response sheet. It reads: If a² + b² + c² = 2(a + c − 1), then the value of a³ + b³ + c³ = ?

Key facts

Over the real numbers a sum of squares is zero only when every square in it is zero.
a² + b² + c² = 2(a + c − 1) rearranges to (a − 1)² + b² + (c − 1)² = 0.
The only real solution is a = 1, b = 0, c = 1, whose cubes total 2.

Study next

Completing the square to pin down variables in a constraint equationThe identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)Equations such as a² + b² + c² = ab + bc + ca, which force a = b = c

Common traps

Reaching for a³ + b³ + c³ = 3abc, which needs a + b + c = 0 — here that sum is 2.
Completing the square on b as well and inventing a (b − 1)² the equation does not contain.
Solving for one variable in terms of the others instead of splitting into three squares.

SSC's algebra block hides a standard identity inside an ordinary-looking equation, and the work is recognising which identity. Quant Q.24 of this same paper does the trigonometric version, where a cosec and cot expression has to be simplified before any value is used.

Related PYQs

No directly related past PYQ was found.