If a = 0.02, b = 0.03, c = −0.05, and a + b + c = 0, find (a³ + b³ + c³) ÷ (3abc)

Correct — C.
Check the condition: 0.02 + 0.03 + (−0.05) = 0
Identity: when a + b + c = 0, a³ + b³ + c³ = 3abc
So the numerator equals the denominator.

Direct check: a³ + b³ + c³ = 0.000008 + 0.000027 − 0.000125 = −0.00009
3abc = 3 × 0.02 × 0.03 × (−0.05) = −0.00009

(a³ + b³ + c³) ÷ (3abc) = −0.00009 ÷ −0.00009 = 1 → option (c)

Why the others are wrong

• (a)
  −1 — −1 lets the negative c flip the sign of the answer. But c makes both parts negative, −0.00009 on top and −0.00009 below, so the quotient is +1.
• (b)
  0 — 0 puts the condition in place of the result. It is a + b + c that is 0, while a³ + b³ + c³ is −0.00009, equal to 3abc.
• (d)
  2 — 2 would need a³ + b³ + c³ = 6abc. The identity gives exactly 3abc, the same as the denominator, so the quotient is 1.

The identity behind this 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 exactly. The ratio asked for is then 1 for any such values, as long as none of a, b, c is 0.

The decimals make direct cubing slow. Checking the sum first replaces all of it with one line.

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.
• If a + b = c, then a³ + b³ − c³ = −3abc.

Common traps

• Cubing 0.02, 0.03 and −0.05 directly and slipping on the decimal places, when the condition settles it
• Choosing −1 because c is negative, without noticing that the numerator and the denominator are both negative

No directly related past PYQ was found.