Question

Show that n^3 - n is divisible by 8 if n is an odd positive integer

Answer 1

Step-by-step explanation:

if n is a odd positive.

then,

n= 2x-1; where x is an odd.

then,

n^3 -n= (2x-1)^3 - (2x-1);

= (2x-1){(2x-1)^2 - 1};

= (2x-1){(2x-1+1)(2x-1-1)}; \\ using a^2 - b^2 =(a+b)(a-b)

= (2x-1)(2x){2(x-1)}

here, (2x-1) is an odd part.

and 2x is an even part as x is an odd.

2x is divisible by 2;

and (x-1) is an even part as x is an odd.

(x-1) divisible by 2.

then n^3-n = (2x-1)(2x){2(x-1)} is divisible by

2*2*2=8.