16.10

$\begin{aligned} S & = \sum_{n=\color{#D61F06}{0}}^{2016} \left \lfloor \frac {n^3}{n^2+1} \right \rfloor & \small \color{#D61F06}{\text{The }0^{th} \text{ term }=0} \\ & = \sum_{n=\color{#D61F06}{1}}^{2016} \left \lfloor \frac {n^3}{n^2+1} \right \rfloor \\ & = \sum_{n=1}^{2016} \left \lfloor \frac {n(n^2+1)-n}{n^2+1} \right \rfloor \\ & = \sum_{n=1}^{2016} \left \lfloor n - \color{#3D99F6}{\frac n{n^2+1}} \right \rfloor & \small \color{#3D99F6}{\text{Note that }0 < \frac n{n^2+1} < 1 } \\ & = \sum_{n=1}^{2016} (n-1) \\ & = \frac {2016(2017)}2 - 2016 \\ & = \boxed{2031120} \end{aligned}$

$\displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n^3}{n^2+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n^3+n-n}{n^2+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor \dfrac{n(n^2+1)-n}{n^2+1}\right\rfloor \implies \displaystyle\sum_{n=0}^{2016}\left\lfloor n- \dfrac{n}{n^2+1}\right\rfloor \\ = \displaystyle\sum_{n=0}^{2016}n-\displaystyle\sum_{n=0}^{2016}\left\lfloor -\dfrac{n}{n^2+1}\right\rfloor \\ = \dfrac{2016\cdot2017}{2}-2016 \quad \quad \left( n<n^2+1 \implies \dfrac{n}{n^2+1}<1 \right) \\ = \boxed{2031120}$