If a + b = c, then the expression a³ + b³ − c³ + 3abc will be equal to:

Correct — A.

The stem is printed as an image: if a + b = c, find a³ + b³ − c³ + 3abc.

Cube the given relation: (a + b)³ = c³
a³ + b³ + 3ab(a + b) = c³
Since a + b = c, that is a³ + b³ + 3abc = c³
So a³ + b³ − c³ = −3abc

Substitute back: (−3abc) + 3abc = 0 → option (a).

Numerical check with a = 1, b = 2, c = 3: 1 + 8 − 27 + 18 = 0.

Why the others are wrong

• (b)
  1 — Substitute a = 1, b = 2, c = 3: 1 + 8 − 27 + 18 = 0, not 1. The expression vanishes for every triple with a + b = c, so no non-zero constant can survive.
• (c)
  3 — 3 is a coefficient, not a value. It sits inside the identity a³ + b³ + c³ − 3abc, and inside the 3abc term here. Test a = 1, b = 1, c = 2: 1 + 1 − 8 + 6 = 0.
• (d)
  2 — a + b − c is a factor of the whole expression, so it dies whenever a + b = c. Try a = 2, b = 3, c = 5: 8 + 27 − 125 + 90 = 0.

Two standard identities settle this, and either is enough.

The first is (a + b)³ = a³ + b³ + 3ab(a + b). Put a + b = c and it reads c³ = a³ + b³ + 3abc, so a³ + b³ − c³ is exactly −3abc and the +3abc in the question cancels it.

The second is quicker once seen. In a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), replace c by −c: the left side becomes the expression asked about, and the first bracket becomes a + b − c, which the condition makes zero.

Key facts

• (a + b)³ = a³ + b³ + 3ab(a + b).
• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca).
• If a + b = c then a³ + b³ − c³ = −3abc, so a³ + b³ − c³ + 3abc = 0 for every such triple.

Common traps

• Expanding term by term instead of using a + b = c on the cube.
• Getting the sign wrong and writing a³ + b³ − c³ = +3abc, which turns the cancellation into 6abc.
• Assuming the answer must involve a, b and c and so rejecting the constant 0.

No directly related past PYQ was found.