If a + b + c = 0 and a = 0.1, b = 0.2, c = −0.3, then what is (a³ + b³ + c³) ÷ (3abc)?

Correct — C.
Check the condition: 0.1 + 0.2 + (−0.3) = 0
Identity: if a + b + c = 0, then a³ + b³ + c³ = 3abc
So (a³ + b³ + c³) ÷ (3abc) = 3abc ÷ 3abc = 1

Verify the numerator: 0.001 + 0.008 − 0.027 = −0.018
Verify the denominator: 3 × 0.1 × 0.2 × (−0.3) = −0.018
Divide: −0.018 ÷ −0.018 = 1 → option (c)

Why the others are wrong

• (a)
  -3 — The signs cancel. Numerator and denominator are both −0.018, so the ratio is +1. A negative answer would need them to have opposite signs.
• (b)
  0 — 0 is the value of a + b + c, not of the ratio. The identity makes the numerator 3abc = −0.018, not 0, so the quotient is 1.
• (d)
  -2 — −2 cannot arise: the identity makes the numerator exactly equal to the denominator 3abc, so the ratio is 1 whenever a + b + c = 0 and abc ≠ 0.

The full identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).

When a + b + c = 0 the right side is 0, so a³ + b³ + c³ = 3abc. The ratio asked here is then 1 for any three numbers that sum to zero, provided none of them is zero.

Cubing the decimals works too, but checking the sum takes one line and settles it.

Key facts

• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
• If a + b + c = 0, then a³ + b³ + c³ = 3abc
• (x − y) + (y − z) + (z − x) = 0, so (x − y)³ + (y − z)³ + (z − x)³ = 3(x − y)(y − z)(z − x)

Common traps

• Answering 0 because a + b + c = 0, when it is a³ + b³ + c³ − 3abc that vanishes
• Losing the minus sign when cubing c: (−0.3)³ = −0.027

No directly related past PYQ was found.