Combinatorial proof for $ \sum _{r=1} ^n r^3 \binom nr = n^2(n+3) 2^{n-3}$ 
Find the combinatorial proof for $$ \sum _{r=1} ^n r^3 \binom nr = n^2(n+3) 2^{n-3}$$

After proving it using algebra, I'm unable to find a combinatorial argument for the above statement. Help would be appreciated! 

Identity $\boldsymbol{1}$: Consider that $r^3$ is the number of ways to pick $3$ items from a set of $r$, with replacement. Break that down into how many duplicates there are.
$0$ duplicates: there are $\binom{r}{3}$ choices of items and $3!=6$ choices of order. $\bbox[5px,border:2px solid #C00000]{6\binom{r}{3}\text{ ways}}$
$1$ duplicate: there are $\binom{r}{2}$ choices of items, $\binom{2}{1}=2$ choices of which is the duplicate and $\binom{3}{2}=3$ choices of order. $\bbox[5px,border:2px solid #C00000]{6\binom{r}{2}\text{ ways}}$
$2$ duplicates: there are $\binom{r}{1}$ choices of the item. $\bbox[5px,border:2px solid #C00000]{\binom{r}{1}\text{ ways}}$
Thus, we have shown
$$
r^3=6\binom{r}{3}+6\binom{r}{2}+\binom{r}{1}\tag{1}
$$
Identity $\boldsymbol{2}$: Consider the number of ways to choose a team of unspecified size with $3$ leaders from a group of $n$ people.
There are $\binom{n}{3}$ choices for the leaders and $2^{n-3}$ ways to choose the remainder of the team. $\bbox[5px,border:2px solid #C00000]{\binom{n}{3}2^{n-3}\text{ ways}}$
Alternatively, for each $r$, there are $\binom{n}{r}$ ways to choose the team and then $\binom{r}{3}$ ways to choose the leaders from the team. $\bbox[5px,border:2px solid #C00000]{\sum\limits_{r=0}^n\binom{n}{r}\binom{r}{3}\text{ ways}}$
Thus, we have shown
$$
\sum\limits_{r=0}^n\binom{n}{r}\binom{r}{3}=\binom{n}{3}2^{n-3}\tag{2}
$$
Conclusion: Putting all this together, we get
$$
\begin{align}
\sum_{r=0}^nr^3\binom{n}{r}
&=6\sum_{r=0}^n\binom{r}{3}\binom{n}{r}+6\sum_{r=0}^n\binom{r}{2}\binom{n}{r}+\sum_{r=0}^n\binom{r}{1}\binom{n}{r}\\
&=6\binom{n}{3}2^{n-3}+6\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\[8pt]
&=2^{n-3}\left(n^3+3n^2\right)\tag{3}
\end{align}
$$
