Square of a sum and of a difference

Step 1 (algebraic proof by direct expansion). Using the distributive law twice, $(a+b)^2=(a+b)(a+b)=a\cdot a+a\cdot b+b\cdot a+b\cdot b=a^2+2ab+b^2$, since $a\cdot b$ and $b\cdot a$ are the same product counted twice.

Step 2 (geometric proof by area). Draw a square of side length $a+b$. Cut it with one horizontal and one vertical line at distance $a$ from one corner. This splits the big square into four pieces: a square of side $a$ (area $a^2$), a square of side $b$ (area $b^2$), and two rectangles of dimensions $a\times b$ (area $ab$ each). Adding the four areas gives $a^2+2ab+b^2$, and since these four pieces exactly tile the original square, this sum must equal $(a+b)^2$.

Step 3 (the difference-of-squares version, by substitution). Replacing $b$ with $-b$ in the first identity gives $(a-b)^2=(a+(-b))^2=a^2+2a(-b)+(-b)^2=a^2-2ab+b^2$, so no separate geometric picture is even needed — the algebra transfers the result automatically.