jmc

Let $a=2013$. The expression is equal to $\frac{a^3-2a^2(a+1)+3a(a+1{)}^2-(a+1{)}^3+1}{a(a+1)}$. We notice a common factor of $a(a+1)$ in the second and third terms of the numerator, so we split the fraction apart: $$\frac{-2a^2(a+1)+3a(a+1{)}^2}{a(a+1)}+\frac{a^3-(a+1{)}^3+1}{a(a+1)}$$The first part of the expression is equal to $-2a+3(a+1)$, which simplifies to $a+3$.

The second part of the expression has a numerator that can be factored as a difference of cubes. We will concentrate on the numerator for now: $$\begin{array}{rl}{a}^3-(a+1{)}^3+1& =(a-(a+1))(a^2+a(a+1)+(a+1{)}^2)+1\\ & =(-1)(a^2+a^2+a+a^2+2a+1)+1\\ & =(-1)(3a^2+3a+1)+1\\ & =-3(a^2+a)-1+1\\ & =-3a(a+1)\end{array}$$Alternatively, we can expand the numerator, and factor: $$\begin{array}{rl}{a}^3-(a+1{)}^3+1& =a^3-(a+1)(a+1)(a+1)+1\\ & =a^3-(a^2+2a+1)(a+1)+1\\ & =a^3-(a^3+3a^2+3a+1)+1\\ & =a^3-a^3-3a^2-3a-1+1\\ & =-3a^2-3a\\ & =-3(a^2+a)\\ & =-3a(a+1).\end{array}$$Taken with the denominator, the second half of the expression simply equals $\frac{-3a(a+1)}{a(a+1)}=-3$.

Putting it back with the first half of the expression, the final expression equals $(a+3)+(-3)=a=\boxed{2013}$.