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! 
 A: From a group of $n$ parliament members, in how many ways can the prime minister choose a cabinet with a finance minister, a home minister and a defence minister? A minister can handle more than one portfolio. 
The cabinet (excluding the prime minister) can be of size $r$ for some $1 \le r \le n$. 
First choose the $r$ ministers, which can be done in $\binom{n}{r}$ ways, then assign the three portfolios; each portfolio can be assigned to any of the $r$ chosen ministers; hence the three portfolios can be assigned in $r \times r \times r = r^3$ ways. So the number of ways of forming the cabinet of $r$ ministers with required portfolio allocations is $r^3 \binom{n}{r}$. Now, $r$, the size of the cabinet can vary from $1$ to $n$. Therefore, the total number of ways of cabinet formation is $\sum_{r=1} ^{n}r^3 \binom{n}{r}$.
Or, the prime minister can assign the three portfolios first, then of the remaining members, select any number of them to the cabinet. There are $5$ disjoint cases.
Case 1:
The three portfolios are assigned to three different members:
 The finance portfolio can be assigned to any of the $n$ members, then the home portfolio can be assigned to any of the remaining $n-1$ members; after that defence can go to any of the remaining $n-2$. Once this is done the prime minister may or may not include each of the remaining $n-3$ members in the cabinet; so each one has $2$ possibilities; either he gets a cabinet berth or he does  not. Therefore, the total number of ways of cabinet formation is $n(n-1)(n-2)2^{n-3}$.
Case 2: The same person gets finance and home; another one gets defence.
 Finance (and home minister) can be chosen in $n$ ways and then defence minister can be chosen in $n-1$ ways. Each of the remaining $n-2$ can be either chosen or not; so each has $2$ possibilities. Number of ways of cabinet formation in this case is $n(n-1)2^{n-2}$.
Case 3: The same person gets finance and defence; another one gets home.
Case 4: The same person gets defence and home; another one gets finance.
Cases 3 and 4 are identical to case 2. Cases 2, 3 and 4 contribute $n(n-1)2^{n-2} + n(n-1)2^{n-2} + n(n-1)2^{n-2} = 3 \times n(n-1)2^{n-2}$ ways of cabinet formation.
Case 5: The same person gets all three portfolios.
This particular minister can be chosen in $n$ ways. Each of the remaining $n-1$ members can be either chosen or not. So number of ways is $n2^{n-1}$.
Thus number of ways of cabinet formation $=n(n-1)(n-2)2^{n-3}+3 \times  n(n-1)2^{n-2} + n2^{n-1}$, this on simplification becomes $n^2 (n+3)2^{n-3}$.
A: 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
$$
\begin{align}
\sum_{r=0}^nr^3\binom{n}{r}
&=6\sum_{r=0}^n\binom{r}{3}\binom{n}{r}+6\sum_{r=0}^n\binom{r}{2}\binom{n}{r}+\sum_{r=0}^n\binom{r}{1}\binom{n}{r}\\
&=6\binom{n}{3}2^{n-3}+6\binom{n}{2}2^{n-2}+\binom{n}{1}2^{n-1}\\[8pt]
&=2^{n-3}\left(n^3+3n^2\right)\tag{3}
\end{align}
$$
