# Cost of LU decomposition (time cost)

After calculation of the cost of the steps of the LU decomposition, and we come to the end result:

$(2/3)n^3 - (2/3)n$ and we say the total cost is then $(2/3)n^3$ (ignoring the term $(-2/3)n$), could we conclude (or say) that the cost of the LU decomposition = $O(n^3)$? or is it $O( (2/3)n^3)$ and we HAVE to write the cost = $O( (2/3)n^3 )$?

You ignore the constant in big-O notation. It would be $O(n^3)$.