Sum and difference of two cubes

Step 1 (algebraic proof by expanding the right side). Expand $(a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3$. The middle terms $a^2b$ and $-a^2b$ cancel, as do $ab^2$ and $-ab^2$, leaving exactly $a^3-b^3$.

Step 2 (geometric proof by volume). Take a cube of edge length $a$ (volume $a^3$) and remove a smaller cube of edge length $b$ from one corner (volume $b^3$); the leftover solid has volume $a^3-b^3$. This leftover solid can be sliced into three rectangular slabs, each of thickness $a-b$: a slab $a\times a\times(a-b)$, a slab $a\times b\times(a-b)$, and a slab $b\times b\times(a-b)$. Their volumes add to $(a-b)a^2+(a-b)ab+(a-b)b^2=(a-b)(a^2+ab+b^2)$, matching the leftover volume exactly.

Step 3 (the sum-of-cubes version, by substitution). Replacing $b$ with $-b$ in the difference-of-cubes identity gives $a^3-(-b)^3=(a-(-b))(a^2+a(-b)+(-b)^2)$, which simplifies to $a^3+b^3=(a+b)(a^2-ab+b^2)$.