If the sum of three numbers is 18 and the sum of their squares is 36, then find the difference between the sum of their cubes and three times of their product.
- (a)1449
- (b)−1494
- (c)4149
- (d)−1944
Answer
Why
Correct — D. The expression asked for is one side of a standard factorisation, so build its two brackets.
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
First bracket = 18, given directly.
For the second bracket, square the sum:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
18² = 36 + 2(ab + bc + ca), so 324 − 36 = 288
ab + bc + ca = 288 ÷ 2 = 144
Second bracket = 36 − 144 = −108
Answer = 18 × (−108) = −1944 → option (d).
Why the others are wrong
- (a)1449 — The sign carries this question. The second bracket is 36 − 144 = −108, a negative, so 18 × (−108) cannot come out positive. 1449 also scrambles the digits of 1944.
- (b)−1494 — −1494 has the right sign and the wrong product. 18 × 108 = 1944, and 18 × 100 is already 1800, so anything under 1800 fails the multiplication before the sign is even checked.
- (c)4149 — 4149 is positive and far too large. 18 × 108 fixes the size at 1944 and the negative bracket fixes the sign, so no route through this data reaches 4149.
Concept
Two identities do all the work here, and neither of them needs a, b or c individually.
The factorisation a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) turns a cubic target into a product of two things you can already evaluate.
The square of a sum, (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), is the bridge: it converts a sum and a sum of squares into the pairwise product term the first identity wants.
The special case worth memorising is a + b + c = 0, where the whole left side collapses to a³ + b³ + c³ = 3abc.
The data cannot be met by real numbers. For any real a, b and c, (a + b + c)² ≤ 3(a² + b² + c²), and this stem would demand 324 ≤ 108.
That does not break the question. Both identities are algebraic and hold regardless, so the intended route still produces a single value. Work the identity and move on rather than hunting for three numbers that do not exist.
Key facts
- a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
- (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), which is how a sum and a sum of squares yield ab + bc + ca.
- When a + b + c = 0 the identity collapses to a³ + b³ + c³ = 3abc.
- For real numbers (a + b + c)² is never more than 3(a² + b² + c²).
Study next
Common traps
- Writing the second bracket as a² + b² + c² + ab + bc + ca and losing the minus signs.
- Reading 324 − 36 = 288 as ab + bc + ca instead of halving it to 144.
- Reporting 1944 as positive because the arithmetic was done on magnitudes.
Nothing here asks for a, b or c. Read the target expression first, pick the identity that produces it, then feed that identity the given sum and sum of squares.
The given pair varies — a sum with a sum of squares, or a sum with a product — while the identity chain stays the same.
Related PYQs
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