27³ + 10³−29³ + 246 is equal to:
- (a)0
- (b)−3460
- (c)−2860
- (d)1
Answer
Why
Correct — B.
Shortcut check: 27 + 10 − 29 = 8, not 0, so a³ + b³ + c³ = 3abc does not apply.
Pair the big cubes with a³ − b³ = (a − b)(a² + ab + b²):
29³ − 27³ = 2 × (841 + 783 + 729) = 2 × 2353 = 4706
Rewrite: 27³ + 10³ − 29³ = 1000 − 4706 = −3706
Add 246: −3706 + 246 = −3460 → option (b)
Why the others are wrong
- (a)0 — 0 assumes the cubes cancel. They do not: 27³ + 10³ = 20683, well short of 29³ = 24389, so the total is negative.
- (c)−2860 — −2860 is 600 above the true total. Recheck the largest cube, where a slip costs most: 29² = 841, and 841 × 29 = 24389.
- (d)1 — 1 has no route from these numbers. 27³ + 10³ + 246 = 20929, and taking away 29³ = 24389 leaves −3460.
Concept
For a sum of cubes, test the sum of the bases first. If a + b + c = 0, then a³ + b³ + c³ = 3abc and no cubing is needed. Here 27 + 10 + (−29) = 8, so that route is closed.
The next-fastest route is the difference of cubes, a³ − b³ = (a − b)(a² + ab + b²). With 29 and 27 just 2 apart, it turns two five-digit cubes into one short multiplication.
The full identity reaches the same total: a³ + b³ + c³ = 3abc + (a + b + c)(a² + b² + c² − ab − bc − ca).
= −23490 + 8 × 2473 = −3706, and adding 246 gives −3460.
Key facts
- a³ + b³ + c³ = 3abc whenever a + b + c = 0.
- a³ − b³ = (a − b)(a² + ab + b²).
- 27³ = 19683 and 29³ = 24389.
Study next
Common traps
- Applying a³ + b³ + c³ = 3abc without first checking that the bases add to zero
- Mis-cubing the largest base: an error in 29³ shifts the total by the same amount
12 Sep 2025, 09:00, Quant Q.25 has the same shape, 31³ + 18³ − 37³ + 210, where 31 + 18 − 37 = 12 and direct working gives the keyed −14820.
17 Sep 2025, 16:00, Quant Q.21 does allow the shortcut: 19 + 20 − 39 = 0, so 19³ + 20³ − 39³ = 3 × 19 × 20 × (−39) = −44460, and adding 118 gives the keyed −44342.
Related PYQs
No directly related past PYQ was found.