Question

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Answer 1

Answer:

8. n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2 [ ∵ if P = ab + r , then 0 ≤ r < a by Euclid lemma ]

∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer

Case 1 :- when n = 3r

Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ]

Case2 :- when n = 3r + 1

e.g., n - 1 = 3r +1 - 1 = 3r

Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3

Case 3:- when n = 3r - 1

e.g., n + 1 = 3r - 1 + 1 = 3r

Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3

From above explanation we observed n³ - n is divisible by 3 , where n is any positive integers

Answer 2

Answer:

8. n³ - n = n(n²-1) = n(n-1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 2 [if P = ab +r, then Osr<a by Euclid lemma]

Let n = 3r, 3r +1, 3r+2, where r is an integer

Case 1 :- when n = 3r

Then, n³ - n is divisible by 3 ["n³-n=

n(n-1)(n+1) = 3r (3r-1)(3r+1). clearly shown it

is divisible by 3]

Case2 :- when n = 3r +1

e.g., n-1=3r +1-1= 3r

Then, n³ - n = (3r+ 1)(3r)(3r+2), it is divisible by 3

Case 3:- when n = 3r - 1

e.g., n+1=3r-1+1=3r

Then, n³ - n = (3r-1)(3r-2)(3r), it is

divisible by 3

From above explanation we observed n³-n is divisible by 3, where n is any positive integers