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  • Prove by direct method that for any integer n, n^{3} - n is always even.

Prove by direct method that for any integer n, n^{3} - n is always even.

Answers (1)

Given : n^{3}-n

Let us consider that, n is even

Thus, n=2k .......(k-natural\; no)

\\n^{3} - n = (2k)^{3} - (2k) \\n^{3} - n = 2k (4k^{2} - 1)

[Let us consider that k(4k^{2}-1)=m]

Thus, n^{3}-n=2m

Thus,

n^{3}-n is even

Now, we will assume that n is an odd no.

Thus, n = (2k + 1) ....... (k - natural \; no.)

\\n^{3} - n = (2k + 1)^{3} - (2k + 1)\\ n^{3} - n = (2k + 1) [(4k2 + 4k + 1 - 1)]

\\n^{3} - n = (2k + 1) [(4k2 + 4k)]\\ n^{3} - n = 4k (2k + 1) (k + 1)

\\n^{2} - n = 2.2k (2k + 1) (k + 1)\\ n^{3} - n = 2\lambda

Thus, from this we can say that n^{3}-n is always even.

 

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