can somebody prove that (k+1)^3 -k^3 is not a cube! That will help me solve this prob.
because 6*n^2 + 12*n + 8 = n^3 + 6*n^2 + 12*n + 8 - n^3 = (n+2)^3 - n^3 = 2((n+2)^2 + n(n+2) + n^2). Now substituting n = 2k We get 2^3 * ( 3*k^2 + 3*k +1). for this to be a cube 3*k^2 + 3*k +1 must be a cube!
3*k^2 + 3*k +1 = (k+1)^3 - k^3.
I am stuck here. All I have to prove is that (k+1)^3 - k^3 is not the cube of a natural number. Any help?