You look at (a + b)^2 and the eye slides to a^2 + b^2. It seems fair: the exponent 2 applies to the whole bracket, and multiplication distributes over addition, so why not exponentiation?

The two expressions are not equal. The correct expansion is:

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

The missing term is 2ab, and it is not a small correction. For a = 3 and b = 4, (3 + 4)^2 = 49, but 3^2 + 4^2 = 9 + 16 = 25. The gap is 24, which is exactly 2ab = 2 \cdot 3 \cdot 4.

The question is why exponentiation does not distribute. The answer is that exponentiation is not multiplication — it is repeated multiplication — and the "distributive" law for multiplication does not extend to iterated multiplication.

The area picture — where 2ab comes from

Here is the argument that should make the identity obvious and unforgettable.

Draw a square with side (a + b). Its area is (a + b)^2 — that is what "squared" means for a length. Now slice the square into four pieces by running one vertical line and one horizontal line at position a along the sides. The four pieces are:

The total area of the big square is the sum of the four pieces:

(a + b)^2 = a^2 + b^2 + ab + ab = a^2 + 2ab + b^2
The square of a plus b decomposed into four areas A large square of side a plus b divided by one vertical and one horizontal line into four regions. The top left is a smaller square of side a, labelled a squared. The top right is a rectangle with sides a and b, labelled a b. The bottom left is a rectangle with sides b and a, labelled a b. The bottom right is a square of side b, labelled b squared. The side lengths a and b are marked on the left and top edges. a² ab ab b² a b a b
A square of side $a + b$ tiled into four pieces. Two diagonal corners hold $a^2$ and $b^2$; the two remaining corners hold identical rectangles of area $ab$ each. The total area is $a^2 + 2ab + b^2$. If you had written $(a+b)^2 = a^2 + b^2$, you would have been measuring only the two coloured squares on the diagonal — leaving the two $ab$ rectangles out of the picture entirely.

The 2ab term is not a correction. It is the area of the two rectangles that have to exist when you draw the square, because the corners of an (a+b) \times (a+b) square are only a^2 and b^2 — the edges between them are where the ab rectangles live.

The algebra, done slowly

If geometry is not your thing, here is the purely symbolic argument. (a + b)^2 is (a + b) multiplied by itself:

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

Use the distributive law on the second bracket:

(a + b)(a + b) = (a + b) \cdot a + (a + b) \cdot b

Why the distributive law applies here: multiplication does distribute over addition. That is the rule x(y+z) = xy + xz, which is one of the axioms from Operations and Properties. The claim that exponentiation distributes is a different (and false) claim.

Now expand each term with the distributive law again:

(a + b) \cdot a = a^2 + ba \quad\text{and}\quad (a + b) \cdot b = ab + b^2

Put the four pieces together:

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

The two middle terms are ba and ab, which are equal (multiplication is commutative), and they sum to 2ab.

Why the "distribute the exponent" instinct is wrong

The rule that sometimes gets called "the distributive law for multiplication" is:

x(y + z) = xy + xz

This multiplies the outside factor into each term of the sum. The analogous-looking rule

(y + z)^n \stackrel{?}{=} y^n + z^n

would exponentiate each term of the sum. It sounds like the same pattern but it is not. The actual rules for exponents are:

The reason for the asymmetry is that (xy)^n is a product of n copies of xy, and the xs and ys can be sorted (commutativity) into a block of xs and a block of ys, giving x^n y^n. But (y+z)^n is a product of n copies of (y+z) — a sum, not a product — and you cannot sort sums apart. Expanding the product creates all possible ways to pick "either y or z" from each of the n copies, which is why cross-terms appear.

The pattern for higher exponents

The identity for the square is the easy one. The cube has more cross-terms, and the fourth power has more still.

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

The coefficients (1, 2, 1), (1, 3, 3, 1), (1, 4, 6, 4, 1) are the rows of Pascal's triangle. The Binomial Theorem is the generalisation of (a+b)^2 = a^2 + 2ab + b^2 to arbitrary exponents:

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k

The only case where the sum collapses to a^n + b^n is the trivial n = 1:

(a + b)^1 = a^1 + b^1

For every higher integer exponent, there are cross-terms. The cross-terms are the default. Zero cross-terms is the exception, and it only happens at the exponent 1.

A numerical sanity check

Pick concrete numbers and verify the identity:

Notice the pattern: the "wrong" answer a^2 + b^2 always undercounts the true (a+b)^2 by exactly 2ab. For a = b, the undercount equals 2a^2, which is a huge fraction of the total.

The reflex

Whenever you see (\text{sum})^n, never collapse the exponent into the terms. Either:

The one thing you may not do is write (a+b)^2 = a^2 + b^2. It is always wrong (for a, b \ne 0). The identity (a+b)^2 = a^2 + 2ab + b^2 is one of the three algebraic identities that you should have in your head as reflex — right next to (a-b)^2 = a^2 - 2ab + b^2 and (a+b)(a-b) = a^2 - b^2.

Related: Exponents and Powers · Tile-View Proof of the Three Core Exponent Laws · Is 2⁻³ Really Negative Eight? The Sign-on-the-Answer Trap · Why a⁰ = 1: The Halving Staircase That Forces the Answer