User:Kamwahchan

The expression (n − 1)3 + (3n − 3)n + 1 for the cube of an integer n is easily reverted because: (n -1)3 + (3n − 3)n + 1= (n3-3n2+3n-1) + (3n2-3n) +1= n3 The following is a brand new expression for expansion of n3 for any positive integer: n3= n + 6 {(n-1)+(n-2)+(n-3)+…[n-(n-1)]+(n-2)+(n-3)+…[n-(n-1)]+(n-3)+(n-4)+ …[n-(n-1)]+(n-4)(n-5)+ …[n-(n-1)]+……[n-(n-1)]} Simplified: n3= n+6{(n-1) + 2(n-2) + 3(n-3) + 4(n-4) +……(n-1)[n-(n-1)]} It is amazing that the cube of any positive integers equals to the sum of multiples of 6 and its cubic root. Consider geometrically, the cube n3 consists of 6{(n-1) + 2(n-2) + 3(n-3) + 4(n-4) +……(n-1) [n-(n-1)]} numbers of the single unit plus n numbers of such units. Where does the wonderful n units hide up inside the cube ?--Kamwahchan (talk) 03:46, 9 August 2010 (UTC)