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

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

Prerequisites:
Distributivity property of multiplication

Proof:
Applying distributivity property of multiplication repeatedly:

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

Hence the result


Recommended:
$ (a + b)(a - b) = a^2 - b^2$
$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
Quadratic Formula