31³ + 18³−37³ + 210 is equal to:
- (a)−36810
- (b)−14820
- (c)−45670
- (d)−23450
Answer
Why
Correct — B. Cube each number, then combine.
31³ = 961 × 31 = 29791
18³ = 324 × 18 = 5832
37³ = 1369 × 37 = 50653
Add the two positive cubes: 29791 + 5832 = 35623
Subtract 37³: 35623 − 50653 = −15030
Add 210: −15030 + 210 = −14820 → option (b)
Why the others are wrong
- (a)−36810 — −36810 is about 2.5 times the true size. 37³ ≈ 50,650 against 31³ + 18³ ≈ 35,620 puts the result near −15,000, nowhere near −36,810.
- (c)−45670 — −45670 is below −44611, the value of 18³ − 37³ + 210 with 31³ left out. Adding the positive 31³ can only raise the total, so it cannot sink this low.
- (d)−23450 — −23450 is below −20652, the value of 31³ − 37³ + 210 with 18³ left out. Adding the positive 18³ = 5832 can only raise the total above that.
Concept
This is direct computation with cubes of two-digit numbers. Building each cube as square × number (37² = 1369, then × 37) keeps it manageable.
Estimate before you multiply: 37³ ≈ 50,650 and 31³ + 18³ ≈ 35,620, so the answer is near −15,000. That alone separates the options.
The shortcut a³ + b³ + c³ = 3abc applies when a + b + c = 0. Here 31 + 18 − 37 = 12, so it does not apply.
A shorter route pairs the two large cubes with the difference-of-cubes identity:
37³ − 31³ = (37 − 31)(37² + 37 × 31 + 31²)
= 6 × (1369 + 1147 + 961) = 6 × 3477 = 20862
18³ − 20862 + 210 = 5832 − 20862 + 210 = −14820
Key facts
- 31³ = 29791, 18³ = 5832 and 37³ = 50653.
- a³ − b³ = (a − b)(a² + ab + b²).
- If a + b + c = 0, then a³ + b³ + c³ = 3abc.
Study next
Common traps
- Applying a³ + b³ + c³ = 3abc without checking that a + b + c = 0. Here the sum is 12, and the shortcut would give 3 × 31 × 18 × (−37) = −61938, which is wrong.
- Dropping the + 210 at the end, which leaves −15030.
The full identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) is tested on 10 Sep 2024, 09:00, Quant Q.22, where a sum of 18 and a sum of squares of 36 give −1944.
Related PYQs
No directly related past PYQ was found.