proving the identity $b^3 = 6\binom{b}{3} + 6\binom{b}{2} + b$ for $b \in \mathbb{N}$, $b > 2$, by counting a set in different ways I'm trying to prove that $\forall b \in \mathbb{N}, b>2,$ $b^{3} = 6\binom{b} {3} +6 \binom{b}{2} +b$  without just using algebra. The idea I've thought about is that there are $b^{3}$ ways to choose a triple of numbers from  $\{1,\ldots, b\}$, and there are obviously $b$ ways to choose one element, the issue is I'm not sure how the $\binom{b}{3}$ and $\binom{b}{2}$ terms contribute. Any help would be greatly appreciated.

Slight different approach (Case 2 below)!
Consider $b=3$ for simplicity. 
There are $b^3=27$ ways to form a $3$-digit number with the digits $1,2,3$.
Case 1: All three digits are the same:
$${b\choose 1}={3\choose 1}=3 \Rightarrow 111,222,333.$$
Case 2: Two digits are the same:
$${3\choose 2} \Rightarrow \color{red}{**\times},\color{green}{*\times*},\color{blue}{\times**} \quad \text{(* - same, $\times$ - different)}\\
{3\choose 2}{b\choose 1}={3\choose 2}{3\choose 1} \Rightarrow \color{red}{11\times,22\times,33\times},\color{green}{1\times1,2\times2,3\times3},\color{blue}{\times11,\times22,\times33}\\
{3\choose 2}{b\choose 1}{b-1\choose 1}={3\choose 2}{3\choose 1}{2\choose 1} \Rightarrow \color{red}{112,113,221,223,331,332},\color{green}{\cdots},\color{blue}{\cdots}.$$
Case 3: All three digits are different:
$${b\choose 3}\cdot 3!={3\choose 3}\cdot 3! \Rightarrow 123,132,213,231,312,321.$$
Hence:
$$\begin{align}b^3&={b\choose 3}\cdot 3!+{3\choose 2}{b\choose 1}{b-1\choose 1}+{b\choose 1}=\\
&=6\cdot {b\choose 3}+{3\choose 2}{b\choose 2}{2\choose 1}+{b\choose 1}=\\
&=6\cdot {b\choose 3}+6\cdot {b\choose 2}+b.\end{align}$$
