SSC CGL 2024 Tier-I · 17 September 2024, Shift 2 · Quantitative Aptitude · Question 19 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 a³ + 3a² + 3a = 7, then the value of a² + 2a is:

(a)4
(b)2
(c)3
(d)1

Answer

Why

Correct — C. The stem gives a³ + 3a² + 3a = 7 and asks for a² + 2a. The left side is one term short of a cube, so add 1 to both sides.

a³ + 3a² + 3a + 1 = 7 + 1
(a + 1)³ = 8
a + 1 = 2, so a = 1.

The asked expression hides the same shift, so add 1 to it as well.
a² + 2a + 1 = (a + 1)² = 2² = 4
a² + 2a = 4 − 1 = 3 → option (c).

Why the others are wrong

(a)
4 — 4 is (a + 1)², one step short of the answer. The stem asks for a² + 2a, so the 1 you added to complete the square has to be taken back off.
(b)
2 — 2 is a + 1, the cube root of 8. That is the middle of the working, not its end — the expression asked for is quadratic, not linear.
(d)
1 — 1 is the value of a itself. Substituting it back gives a² + 2a = 1 + 2 = 3, so stopping at a drops the final addition.

Concept

Both expressions in this item are shifted binomials. a³ + 3a² + 3a + 1 is (a + 1)³, and a² + 2a + 1 is (a + 1)².

Whenever a polynomial sits one constant away from a perfect power, add that constant to both sides instead of solving for the variable. A cubic collapses into a cube root in a single line.

The same shift then unlocks the expression you are asked for, which is why you often never need the value of a at all: a² + 2a = (a + 1)² − 1 = 4 − 1 = 3.

Solving for a first also works, because (a + 1)³ = 8 has the single real root a = 1.

But that route is longer, invites arithmetic slips, and stops working the moment the right-hand side is not a perfect cube. Reading a³ + 3a² + 3a as (a + 1)³ − 1 always works.

Key facts

(a + 1)³ = a³ + 3a² + 3a + 1, so a³ + 3a² + 3a = (a + 1)³ − 1.
(a + 1)² = a² + 2a + 1, so a² + 2a = (a + 1)² − 1.
With a³ + 3a² + 3a = 7 the cube becomes (a + 1)³ = 8, giving a + 1 = 2.
8 has exactly one real cube root, so a = 1 is the only real solution here.

Study next

Spotting (x ± 1)³ and (x ± 1)² inside an already expanded polynomialSubstituting a whole bracket, such as y = a + 1, to reduce a cubic to a one-step equationThe paired identities that build x³ + 1⁄x³ and x² + 1⁄x² from x + 1⁄x

Common traps

Reporting (a + 1)² = 4 without subtracting the 1 that was added to complete the square
Reporting the cube root a + 1 = 2 as the value of the expression
Expanding and solving the cubic by trial when adding 1 finishes it in one line

SSC prints this as a one-line stem: one expression is given a value, and a second, related expression is asked for.

The arithmetic is trivial once the shift is spotted, so what the item really tests is whether you recognise the identity rather than whether you can solve a cubic.

Related PYQs

No directly related past PYQ was found.