Visual walkthrough — Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³)

Before we start, two words we must agree on, in plain language:

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Step 1 — Lay down the two sticks

WHAT. We start with two lengths, $a$ and $b$. Nothing else exists yet. In the picture $a$ is the longer lavender stick, $b$ is the shorter coral stick.

WHY. Every identity in the parent note is a statement about two quantities added together. Before we can talk about $(a+b)$ we need to physically have an $a$ and a $b$ to add. Starting from sticks (lengths) rather than symbols keeps us honest — we will never write a symbol we cannot draw.

PICTURE.

$\underset{\text{long stick}}{\underbrace{a}}\underset{\text{short stick}}{\underbrace{b}}$

There is no equation to expand yet — just two labelled lengths sitting on the table.

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Step 2 — Glue the sticks into one length $(a+b)$

WHAT. Put the two sticks end to end. The combined stick has length $a+b$ — read it aloud as "a-plus-b", a single new length.

WHY. The identity is about $(a+b{)}^2$, and the parentheses mean "treat $a+b$ as one thing". Physically joining the sticks makes that "one thing" real: it is a single segment whose total length just happens to be built from an $a$-part and a $b$-part.

PICTURE.

$\underset{\text{first part}}{\underbrace{a}}+\underset{\text{second part}}{\underbrace{b}}=\underset{\text{the whole joined stick}}{\underbrace{(a+b)}}$

• The $+$ is not an instruction to compute — it just records where the join is, so we don't lose track of which part was $a$ and which was $b$.

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Step 3 — Build a square on that joined length

WHAT. Take the joined stick of length $(a+b)$ and build a square on it: a square whose side is $(a+b)$. Its area is, by the very definition of "square of a length", $(a+b{)}^2$.

WHY. Why a square and not, say, a triangle? Because squaring literally means "make the square". The whole quest — $(a+b{)}^2=?$ — is the question "how big is this square?" We are going to answer it by measuring the same square two different ways. If two honest measurements of one shape disagree, we made an arithmetic error; if they agree, we have an identity.

PICTURE.

$\text{side}=(a+b)⟹\text{area}=\underset{\text{the number we want to name}}{\underbrace{(a+b{)}^2}}$

• Notice the side along the bottom and the side going up are the same length $(a+b)$, because it is a square. That sameness is what makes the next cut work.

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Step 4 — Mark the $a$/$b$ join on both sides and slice

WHAT. On the bottom side, mark where the $a$-part ends and the $b$-part begins. Do the identical mark on the left side. Now draw one horizontal line and one vertical line through those marks, straight across the square.

WHY. The two cuts turn one big square into a grid of four smaller rectangles. This is the only clever move on the whole page, and it rests on the "area is preserved when you slice" idea from the top. We haven't added or removed any area — we've just drawn lines. So:

$\underset{\text{big square, one measurement}}{\underbrace{(a+b{)}^2}}=\underset{\text{four small pieces, second measurement}}{\underbrace{\text{(piece)}+\text{(piece)}+\text{(piece)}+\text{(piece)}}}$

PICTURE.

The dashed lavender line and dashed coral line split the square into a $2\times 2$ grid. We name the pieces next.

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Step 5 — Measure each of the four pieces

WHAT. Read off the side lengths of each little rectangle from the grid, and multiply length $\times$ width to get its area.

WHY. Each piece is a plain rectangle, and the area of a rectangle is just (one side) $\times$ (other side) — no identities needed, only multiplication. Watch each factor come directly from a mark on the grid:

$$\begin{array}{ccc}\text{piece}& \text{sides}& \text{area}\\ \text{top-left (lavender)}& a\times a& a^2\\ \text{top-right (mint)}& a\times b& ab\\ \text{bottom-left (mint)}& b\times a& ab\\ \text{bottom-right (coral)}& b\times b& b^2\end{array}$$

• The corner square sits where the $a$-part meets the $a$-part → sides $a$ and $a$ → area $a^2$. It is doing the job of "the square of the first term".
• The two mint rectangles both have one side $a$ and one side $b$. That is why there are two copies of $ab$ — one lives above the cut, one lives to the side of it. This is the visual birthplace of the "$2$" that trips everyone up.
• The little coral square is $b$ by $b$ → area $b^2$, "the square of the second term".

PICTURE.

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Step 6 — Add the pieces and collect the twins

WHAT. The big square's area equals the sum of its four pieces:

$$(a+b{)}^2=\underset{\text{corner square}}{\underbrace{a^2}}+\underset{\text{mint }\#1}{\underbrace{ab}}+\underset{\text{mint }\#2}{\underbrace{ab}}+\underset{\text{little square}}{\underbrace{b^2}}$$

The two mint rectangles are identical in area, so $ab+ab=2ab$:

$$\boxed{(a+b{)}^2=a^2+\underset{\text{the two twins, merged}}{\underbrace{2ab}}+b^2}$$

WHY. We measured one shape in two ways: once as a whole (Step 3), once as four pieces (Step 5). Both are the true area, so they must be equal. That equality is the identity — not a rule handed down, but a forced conclusion.

PICTURE.

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Step 7 — The degenerate cases (nothing is allowed to break)

WHAT. Check the picture still tells the truth when a stick has length zero, or when we use the difference $(a-b)$ instead.

WHY. A good picture must survive its extreme inputs. If our square dissolves sensibly at the edges, we trust it everywhere in between.

Case $b=0$. The coral stick vanishes. The right column and bottom row of the grid shrink to nothing: both mint pieces and the coral square disappear. Only $a^2$ survives: $(a+0{)}^2=a^2+2a(0)+0^2=a^2.✓$

Case $a=b$. Both sticks equal, so all four pieces become the same size. The area is four equal squares: $(a+a{)}^2=a^2+2a\cdot a+a^2=4a^2=(2a{)}^2.✓$

Case difference $(a-b{)}^2$. Now start from the big $a\times a$ square and chop off a strip of width $b$. You remove two $ab$ strips but you subtracted the shared corner twice, so you must add one $b^2$ back: $(a-b{)}^2=a^2-2ab+b^2.$ The lone $+b^2$ is the corner you over-removed — the same visual logic, run in reverse.

PICTURE.

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The one-picture summary

One square, one horizontal cut, one vertical cut, four labelled tiles: $a^2$, two $ab$'s, and $b^2$. Read the whole square = read the four tiles ⇒ $(a+b{)}^2=a^2+2ab+b^2$.

Recall Feynman retelling — say it to a friend with no symbols

I took two sticks, a long one and a short one, and glued them into one longer stick. I built a square on that long stick, so its area is "the long stick squared". Then I marked on every side exactly where the long-part ended and the short-part began, and I cut straight across in both directions. That chopped my square into four floor-tiles: a big one where long meets long, a tiny one where short meets short, and — crucially — two identical middle tiles, each with one long side and one short side. Since I only cut and never added or removed any floor, the big square must equal all four tiles added up: one big + one tiny + two middles. Written in symbols that's $a^2+b^2+2ab$. The whole reason there's a "2" in the answer is that the middle tile shows up twice — once above the cut and once beside it. And if I ever try to say the answer is just $a^2+b^2$, I'm pretending those two middle tiles aren't real, when I can literally see them.

Recall Quick self-test

Where do the two $ab$ terms physically come from? ::: The two identical middle rectangles in the grid — one has sides $a$(across) $\times b$(up), the other $b\times a$; both have area $ab$. Why must the answer have a $+b^2$ even for $(a-b{)}^2$? ::: When you chop a width-$b$ strip off two sides of the $a$-square, the corner $b^2$ gets removed twice, so you add one copy back. Set $a=5,b=2$: what is $(a+b{)}^2$ by tiles? ::: $25+2(10)+4=49=7^2$. ✓

Related tools: the cutting-and-summing move is just the Distributive Property drawn as area; doing it symbolically is the FOIL Method; running it backwards is Factoring Polynomials; the coefficients $1,2,1$ (and $1,3,3,1$ for cubes) are read off Pascal's Triangle via the Binomial Theorem; and adding the missing $b^2$ back is the heart of Completing the Square.