If m + n = 24, then (m − 16)³ + (n − 8)³ is ____.

Correct — A. Nothing needs substituting — the two brackets are negatives of each other.

Let a = m − 16 and b = n − 8.
a + b = (m + n) − 24 = 24 − 24 = 0

Now use a³ + b³ = (a + b)(a² − ab + b²).
The first factor is 0, so the whole product is 0 → option (a)

Test it on any pair: m = 20, n = 4 gives 4³ + (−4)³ = 64 − 64 = 0, and m = 24, n = 0 gives 8³ + (−8)³ = 0.

Why the others are wrong

• (b)
  320 — The expression is 0 for every pair with m + n = 24, so no fixed non-zero value can be right. Try m = 20 and n = 4: you get 4³ + (−4)³ = 0, not 320.
• (c)
  576 — 576 is 24², the square of the sum printed in the stem. That is pattern-matching the numbers on the page, and the identity a³ + b³ = (a + b)(a² − ab + b²) never produces it.
• (d)
  40 — 40 is simply 16 + 24, two numbers lifted straight off the page. The 16 and the 8 are there to make the brackets cancel, not to be added to anything.

The identity is a³ + b³ = (a + b)(a² − ab + b²), and this whole question lives in that first factor.

When a stem hands you a sum of two linear expressions, add them before doing anything else. Here (m − 16) + (n − 8) = m + n − 24, and the stem sets m + n = 24, so the pair sums to zero.

Two numbers that add to zero are negatives of each other, and x³ + (−x)³ = 0 whatever x is.

That is why m and n are never pinned down individually, and never need to be.

Key facts

• a³ + b³ = (a + b)(a² − ab + b²), so a + b = 0 forces the sum of cubes to 0.
• Here (m − 16) + (n − 8) = m + n − 24 = 0.
• The three-variable companion is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), which vanishes when a + b + c = 0.

Common traps

• Trying to solve for m and n, which the stem never determines.
• Expanding both cubes in full and losing the cancellation somewhere in the arithmetic.
• Rejecting 0 on the grounds that it looks too easy for a marked question.

No directly related past PYQ was found.