Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$ 
Why does $(a + b)^3$ expand to $a^3 + 3ab^2 + 3a^2b + b^3$

Why does this work? I am confused as to why this happens. 

Algebra meets combinatorics.  The exponent $3$ tells us to expand the product of three binomials
\begin{align*}
(a+b)^3=(a+b)(a+b)(a+b)
\end{align*}
From each of the three factors $a+b$ we have to select either $a$ or $b$.

Let's list the different possibilities:
  
  
*
  
*From each    binomial   we select an  $a$. Since there is only one way to do so we obtain
  \begin{align*}
aaa\qquad\rightarrow\qquad \color{blue}{1a^3}\qquad\qquad\ \ 
\end{align*}
  
*Number of ways to select two  $a$'s and  one $b$
  \begin{align*}
aab,\ aba,\ baa\qquad\rightarrow\qquad aab+aba+baa=\color{blue}{3a^2b}
\end{align*}
  
*Number of ways to select two  $b$'s and  one $a$
  \begin{align*}
bba,\ bab,\ abb\qquad\rightarrow\qquad bba+bab+abb=\color{blue}{3ab^2}
\end{align*}
  
*Number of ways to select three  $b$'s
  \begin{align*}
bbb\qquad\rightarrow\qquad \color{blue}{1b^3}\qquad\qquad\ \ 
\end{align*}
  
*No  other possibilities.
  
  
  We conclude
  \begin{align*}
(a+b)^3=a^3+3a^2b+3ab^2+b^3
\end{align*}

Hint:  Observe the nice symmetry between $a$ and $b$.
Challenge: Try a similar approach with $(a+b)^\color{blue}{4}$ and verify the result algebraically. This is more complex but still feasible .
