19³ + 20³−39³ + 118 is equal to:

Correct — A. Take a = 19, b = 20, c = −39, so the expression is a³ + b³ + c³ + 118.

Add the bases: 19 + 20 − 39 = 0
When a + b + c = 0, a³ + b³ + c³ = 3abc

Multiply the positives: 3 × 19 × 20 = 1140
Multiply by c = −39: 1140 × (−39) = −44460
Add 118: −44460 + 118 = −44342 → option (a)

Why the others are wrong

• (b)
  −6789 — −6789 is far too small in magnitude. 39³ = 59319 against 19³ + 20³ = 14859 leaves −44460, and adding 118 moves that by only 118.
• (c)
  0 — 0 mixes up the identity: a + b + c = 0 makes the bases sum to zero, not the cubes. The cubes sum to 3abc = −44460.
• (d)
  1 — 1 would need the three cubes to total −117. They total −44460, because 39³ = 59319 outweighs 19³ + 20³ = 14859.

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

When a + b + c = 0 the right side is zero, so a³ + b³ + c³ = 3abc. Spotting 19 + 20 = 39 turns three large cubes into one short product.

Key facts

• If a + b + c = 0, then a³ + b³ + c³ = 3abc.
• 19³ = 6859, 20³ = 8000, 39³ = 59319.
• A negative c keeps its sign inside 3abc, which makes the product negative here.

Common traps

• Answering 0 because a + b + c = 0. The bases sum to zero, the cubes sum to 3abc.
• Dropping the minus sign on 39 and getting 44460 + 118 = 44578.
• Using 3abc without first checking that a + b + c = 0.

No directly related past PYQ was found.