Prove by direct method that for any integer ‘n’, n3 – n is always even.

We have given, n3-n

Let us Assume, n is even

Let n=2k, where k is natural number

n3-n=(2k)3-(2k)

n3-n=2k(4k2-1)

let k(4k2-1)=m

n3-n=2m

Therefore, (n3-n) is even.

Now, Let us Assume n is odd

Let n=(2k+1), where k is natural number

n3-n=(2k+1)3-(2k+1)

n3-n= (2k+1)[(2k+1)2-1]

n3-n= (2k+1)[(4k2+4k+1-1)]

n3-n= (2k+1)[(4k2+4k)]

n3-n= 4k(2k+1)(k+1)

n2-n= 2.2k(2k+1)(k+1)

let λ=2k(2k+1)(k+1)

n3-n=2λ

therefore, n3-n is even.

Hence, n3-n is always even

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