Question

for any positive integer n then (N^3-n)is divisible by
a. 6
b. 8
c. 2
d. 5

Answer 1

Given expression ,

n³-n
= n(n²-1)
= n(n+1)(n-1)

Hence, In a triplet of three consecutive numbers, At least one of the two is even.

So, n³-n is divisible by 2 .

Also, n + n-1 + n+1 = 3n which is divisible by 3 0

Sum of digits is divisible by 3 , Hence, This is divisible by 3 .

So, n³-n is divisible by 2 , 3 .As it is divisible by both 2 , 3 it is also divisible by 6 .

Required options as A, C.