Prove by direct method that for any integer 'n',`n^(3)` - n is always even.

Answer

N/a

Explanation

Here, two cases arise
Case. I When n is even.
Let `" " n = 2K, K in N`
`rArr " "n^(3) = (2K)^(3) - (2K) = 2K (4^(2) -1)`
` " "= 2 lambda,` Where `lambda = K (4K ^(2) -1)`
Thus ,`(n^(3) - n)` is even when n is ever.
Case II When n is odd,
Let `" "n = 2K + 1, K in N`
`rArr" "n^(3) - n=(2K+1)^(3) - (2K +1)`
`" " =(2K +1) [(2K + 1)^(2)-1]`
`" " = (2K + 1) [( 4K^(2) + 1+ 4K -1]`
`" " =(2K + 1)(4K^(2) + 4K)`
`" " = 4K (2K + 1)(K+1)`
`" " 2 mu, "When" mu = 2K (k-1) (2K +1)`
Then, `n^(3)-n` is even n is odd.
So, `n^(3) - n`is always even.