Question

if a and b are natural number s and a-b is divisible by 3 then a3-b3 is divisible by which number

Answer 1

If a − b is divisible by 3, then a − b = 3k, for some integer k
(a − b)² = (3k)²
a² − 2ab + b² = 9k²
a³ − b³ = (a−b) (a² + ab + b²)
. . . . . = (a−b) (a² − 2ab + b² + 3ab)
. . . . . = 3k (9k + 3ab)
. . . . . = 3k * 3 (3k + ab)
. . . . . = 9 k(3k+ab)
Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9
Answer: 9
I hope this will help you