Question

. For any positive integer n , prove that n³ – n divisible by 6.
non in of the form ha + 5. then it is​

Answer 1

Let n=2

then, n^3-n

= 2^3-2

= 8-2

=6

then 6 divide by 6 is

=6÷6

=1

than, 1+5

=6

Answer 2

Answer:

let n be any positive integer

now dividing and by 6 we get q as quotient and r as reminder.

then by euclid's division lemma

• n=6×q+r. 0<r<6 (r=1,2,3,4,5)
• n³-n =n(n²-1)
• n(n+1)(n-1) [(a+b)(a-b)]
• n=6q+0 =6q
• 6q(6q+1) (6q-1)
• n=6q+0=6q (r=0)
• Now, n(n+1)(n-1) = 6q(6q+1)(6q-1)
• 6×m [ where m = q×(6q+1) (6q-1) ]
• Therefore, n³-n is divisible by 6

Case 2:

• when r=1
• Then n. =6q+1
• now, n (n +1) (n-1) = 6q+1 (6q + 1 + 1) (6q + 1 - 1)
• 6q +1 (6q+2) (6q)
• 6×m ( m=q (6q+1) (6q+2)
• n³-n is divisible by 6

In both cases we see, n³-n is divisible by 6.