SSC CGL 2025 Tier-I · 19 September 2025, Shift 2 · Quantitative Aptitude · Question 7 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 13 September 2026.

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If a³ + b³=35 and ab = 6, find the value of a + b.

(a)5
(b)6
(c)7
(d)8

Answer

Why

Correct — A.
Identity: a³ + b³ = (a + b)³ − 3ab(a + b)
Let s = a + b and put in ab = 6: a³ + b³ = s³ − 18s
Set it equal to 35: s³ − 18s = 35

Try s = 5: 5³ − 18 × 5 = 125 − 90 = 35, which fits
Check: a + b = 5 and ab = 6 give a, b = 2 and 3, and 2³ + 3³ = 8 + 27 = 35

So a + b = 5 → option (a)

Why the others are wrong

(b)
6 — s = 6 gives 6³ − 18 × 6 = 216 − 108 = 108, not 35. A sum of 6 with ab = 6 would need a³ + b³ = 108.
(c)
7 — a + b = 7 with ab = 6 means a, b = 1 and 6, and 1³ + 6³ = 217. The formula agrees: 343 − 126 = 217, far above 35.
(d)
8 — s = 8 gives 512 − 144 = 368, more than ten times the given 35. Past s = 5 the value of s³ − 18s only keeps rising, so no larger sum can work.

Concept

The identity (a + b)³ = a³ + b³ + 3ab(a + b) ties together the sum, the product and the sum of cubes. Know any two and the third follows.

Here the sum is the unknown, so the identity turns into a cubic in s = a + b: s³ − 18s = 35.

With four numbers on offer, substituting each option is faster than solving the cubic.

No other real value of a + b fits. The cubic factors as (s − 5)(s² + 5s + 7) = 0, and s² + 5s + 7 has discriminant 25 − 28 = −3, so it has no real root.

Key facts

a³ + b³ = (a + b)³ − 3ab(a + b)
a³ + b³ = (a + b)(a² − ab + b²), which gives 5 × (13 − 6) = 35 here
a = 2, b = 3 gives a + b = 5, ab = 6 and a³ + b³ = 35

Study next

a³ − b³ = (a − b)³ + 3ab(a − b), for items that give a − b and abx³ + 1⁄x³ = (x + 1⁄x)³ − 3(x + 1⁄x), the same identity with ab = 1a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

Common traps

Writing 3ab in place of 3ab(a + b), which gives s³ = 35 + 18 = 53 and no whole-number sum
Getting the sign wrong: the cube of the sum is a³ + b³ plus 3ab(a + b), so a³ + b³ is the cube minus it

17 Sep 2025, 09:00, Quant Q.6 runs the same numbers forwards: a + b = 5 and ab = 6 give a³ + b³ = 125 − 90 = 35, and the item then asks for 35² − 9 × 36 × 25 = −6875.

Quant Q.24 of this paper uses the three-variable identity instead: a + b + c = 0 makes a³ + b³ + c³ = 3abc, so the ratio asked is 1.

Related PYQs

No directly related past PYQ was found.

Topic: Basic algebraic identities of School Algebra & Elementary surds — all 182 SSC CGL questions on it, every solved shift.