$ (a + b)^3 = a^3 + 3a^2b + 3ab^2+b^3$

Theorem: $ (a + b)^3 = a^3 + 3a^2b + 3ab^2+b^3$

Prerequisites:
$ (a + b)^2 = a^2 + 2ab + b^2$ (proof)                                      

Proof:

$ \begin{align}
\qquad\quad LHS &= (a + b)^3\\
&= (a + b)^2(a + b)\end{align}$

Putting $(a + b)^2 = a^2 + 2ab + b^2$ and applying distributivity property of multiplication,

$\begin{align}\qquad\quad LHS &= (a^2 + 2ab + b^2) (a + b)\\
&= (a^2 + 2ab + b^2)a + (a^2 + 2ab + b^2)b\\
&= (a^3 + 2a^2b + ab^2) + (a^2b + 2ab^2 + b^3)\\
&= a^3 + 3a^2b + 3ab^2+b^3\\
&= RHS
\end{align}$

Hence the result


Recommended:
$ (a + b)^2 = a^2 + b^2 + 2ab$
$ (a + b)(a - b) = a^2 - b^2$