Reproduce the proof. When n = 1, we have
n³ + 2n = 1³ + 2 = 3
which is clearly divisible by 3.
Assume n³ + 2n is divisible by 3. Then for the next natural number n + 1, we have
(n + 1)³ + 2 (n + 1) = n³ + 3n² + 3n + 1 + 2n + 2
= n³ + 2n + 3 (n² + n + 1)
The first two terms n³ + 2n are divisible by 3 according to our assumption, and 3 (n² + n + 1) is clearly a multiple of, and therefore divisible by, 3 as well.
So the claim that 3 divides n³ + 2n is true for all natural numbers n.
