Question

for any positive integers 'n' prove that n³-n is divisible by 6 (or )show that the product ​

Answer 1

Answer:

if a number is divided by 6 the remainder can be

0,1,2,3,4,5. So n could be of the form

n = 6q + 0

n= 6q+1

n= 6q+2

n= 6q+3

n= 6q+4

n= 6q+5

n^3 - n for remainder 0

n^3 - n = (6q)^3 - 6q

= 6( (6^2)q^3- q)

=6(36q^3 -q)

As we can see it is divisible by 6 since there is no remainder.

For remainder 1

n^3 - n = (6q+1)^ 3 - 6q- 1

= (6q)^3 + 1^3 + 3(6q)^2(1) + 3(6q)1^2 - 6q- 1

= 216q^3 +1 + 3(36q^2) + 18q -6q -1

= 216q^3 + 108q^2 + 12 q

= 6( 36q^3 + 18q^2 + 2q)

As we can see it is divisble by 6

Do the same for all the cases where remainder is 2,3,4,5