If A And B Are Natural Numbers And A-b Is Divisible By 3, Then A3-b3

1.	If A And B Are Natural Numbers And A-b Is Divisible By 3, Then A3-b3 Is Divisible By?
Answer» 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. 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.

If A And B Are Natural Numbers And A-b Is Divisible By 3, Then A3-b3 Is Divisible By?

Answer»

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.

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.

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