Question

The sum of the cubes of three consecutive natural numbers is divisible by

itti simple si thoerem nhi aati tm log ko ​

Answer 1

Answer:

n3+(n+1)3+(n−1)3

=3(n)3+6n

=3n(n2+2)

=3n(n2−1+3)

=3(n(n−1)(n+1)+3n)

n(n−1)(n+1) and 3n are both divisible by 3. Hence their sum is also divisible by 3. 

Therefore the whole number is divisible by 9.

Answer 2

The sum of the cubes of three consecutive natural numbers is divisible by 9

Refer the attachment for complete answer ✔️