Question

The sum of the cubes of any three consecutive natural numbers is divisible by

• A. \(9\)
• B. \(8\)
• C. \(6\)
• D. \(3\)

Hint:

For “always divisible” questions, use algebraic expansion and factorization instead of testing many values.

Answer 1

The Correct Option is D

Concept: Let the three consecutive natural numbers be: \[ n,\;n+1,\;n+2 \] Their sum of cubes is: \[ n^3+(n+1)^3+(n+2)^3 \] We check divisibility by simplifying.

Step 1: Expand the terms.
\[ n^3+(n^3+3n^2+3n+1)+(n^3+6n^2+12n+8) \] \[ =3n^3+9n^2+15n+9 \]

Step 2: Take common factor.
\[ =3(n^3+3n^2+5n+3) \] Thus the whole expression is always divisible by: \[ 3 \]

Step 3: Check stronger divisibility.
Take: \[ n=1 \] Then: \[ 1^3+2^3+3^3=1+8+27=36 \] Divisible by \(3\), but not always by \(8\) or \(9\). Thus, the guaranteed divisor is: \[ \boxed{3} \]