SSC CGL 2024 Tier-I · 26 September 2024, Shift 3 · Quantitative Aptitude · Question 8 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 10 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 x ⁄ (x² − 2x + 1) = 1⁄3, then the value of x³ + 1⁄x³ is:

(a)110
(b)180
(c)160
(d)130

Answer

Why

Correct — A. Divide through by x to force x + 1⁄x into view.

x ⁄ (x² − 2x + 1) = 1⁄3, so (x² − 2x + 1) ⁄ x = 3
x − 2 + 1⁄x = 3
x + 1⁄x = 5
x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x) = 125 − 15 = 110

That is option (a).

Why the others are wrong

(b)
180 — 180 cannot follow from x + 1⁄x = 5. The identity a³ − 3a returns 110 at a = 5 and 198 at a = 6, so 180 belongs to no whole value of a at all.
(c)
160 — 160 would need a correction of +35 applied to 5³ = 125. The identity fixes that correction at −3a, which is −15, and its sign never changes.
(d)
130 — 130 is 125 + 5 — adding a where the identity subtracts 3a. Expanding (x + 1⁄x)³ gives x³ + 1⁄x³ + 3(x + 1⁄x), so the 3a comes off, not on.

Concept

The denominator x² − 2x + 1 is (x − 1)², but factorising is not the move here. Dividing every term by x is.

That division turns the equation into the one quantity the whole identity family keys on, a = x + 1⁄x:
x² + 1⁄x² = a² − 2
x³ + 1⁄x³ = a³ − 3a

The equation hands you a = 5 in two lines, and the cube follows without ever solving for x.

The stem is printed as an image rather than as text. It reads: If x ⁄ (x² − 2x + 1) = 1⁄3, then the value of x³ + 1⁄x³ is:

You never need x itself. Rearranged, the equation is x² − 5x + 1 = 0, whose roots are (5 ± √21) ⁄ 2, and cubing an irrational root costs a minute you do not have in Tier-I.

Key facts

Dividing x ⁄ (x² − 2x + 1) = 1⁄3 through by x gives x + 1⁄x = 5.
x³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x).
With x + 1⁄x = 5 the value is 125 − 15 = 110.

Study next

The companion identity x³ − 1⁄x³ = (x − 1⁄x)³ + 3(x − 1⁄x)Getting x² + 1⁄x² and x⁴ + 1⁄x⁴ from the same starting valueEquations handed over as x² − 3x + 1 = 0, which want the same division by x

Common traps

Dropping the −2 from the denominator and taking x + 1⁄x as 3.
Solving the quadratic and cubing an irrational root under time pressure.
Adding 3a instead of subtracting it, which is the shape of the x³ − 1⁄x³ identity.

SSC prints this family as a bare equation with no setting or story. This one is arranged so that a single division by x produces x + 1⁄x, and spotting that shape is most of the work — the rest is a memorised identity.

Related PYQs

No directly related past PYQ was found.