1) let n=2 to prove it. (this was true).
2) let n=k --> k^3-k= f(k), f(k) being a multiple of 6 3) let n=k+1 --> (k+1)^3-k+1
4)when expanded and subtracting k+1 --> k^3+3k^2+2k= 6n where n is a multiple of 6
5)substitute f(k) into equation (f(k)+k)+3k^2+2k=f(k)+3k^2+3k=6n
6)simplify -->f(k)+3(k^2+k)=6n