divisibility of cubes

Divisibility of cubes

We know that a³ – a is equal to a product of three successive numbers (a-1),a,and (a+1).

a³ – a = (a-1)a(a+1). Since the product of any three successive numbers is invariably divisible by 6,

(a-1)a(a+1) = 6n

i.e., a³ – a = 6n  or a³ = a+6n or a² = 1 + [6n/a] i.e., 6n is divisible by a

It is noted that there is some kin d of regularity in the quotient and the remainder when a³ is divided by (a+1),(a+2),(a+3)…..

When a cube a³ is divided by (a+1) or (a-1) ,the quotient will be a(a-1) or a(a+1) with remainder a.

When a³ is divided by (a+2)

a=1 ,2(3) – 5

a=2, 3(4) – 4

a=3, 6(5) -3

a=4,11(6) – 2

a=5,18(7) – 1

a=6,27(8) + 0

a=7, 38(9) + 1

a=8,  51(10) +2

a=9, 66(11) +3

In general, it is in the form,

               a-2

a³ = [ 2+ Σ (2n+1)](a+2) + (a-6)

           0

But               n-2

Σ (2n+1)] is a square number.It is equal to square of a number equal tp the

0

  number of odd numbers added together.i.e, (n-1)² .

a³ = [2+(a-1)²](a+2) + (a-6)

It shows that when a³ is divded by (a+2) , the quotient will be [2+(a-1)²] and the remainder will be (a-6)