for any positive integers not,prove that ncube -n is divisible by 6
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n³ - n
= n(n²-1)
= n(n+1)(n-1)
= (n-1)(n)(n+1)
n³-n is the product of three consecutive numbers. It is obvious that at least one of three consecutive numbers is even and divisible by 2 .
Also, n(n+1)(n-1) is divisible by 3 because
n+n+1+n-1 = 3n ( 3n is divisible by 3 :- Divisibility rule of three ) .
As the result of n³-n is divisible by both 2 ,3 it is divisible by 6 as prime factors of 6 are 2 ,3 .
Hence proved!
= n(n²-1)
= n(n+1)(n-1)
= (n-1)(n)(n+1)
n³-n is the product of three consecutive numbers. It is obvious that at least one of three consecutive numbers is even and divisible by 2 .
Also, n(n+1)(n-1) is divisible by 3 because
n+n+1+n-1 = 3n ( 3n is divisible by 3 :- Divisibility rule of three ) .
As the result of n³-n is divisible by both 2 ,3 it is divisible by 6 as prime factors of 6 are 2 ,3 .
Hence proved!
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