Special products: where the middle term comes from

Why is $(a+b)^2$ different from $a^2+b^2$ ? Enlarging both sides of a square tile adds two strips, as well as a corner square. Those strips supply the middle term $2ab$ . We can see the structure, then recover it without a drawing using polynomial multiplication.

Four parts of a squared sum

Take a square of side $a+b$ , with $a>0$ and $b>0$ . Splitting each side separates a square of area $a^2$ , two strips of area $ab$ , and a small square of area $b^2$ . Adding only the squares would leave the strips uncounted.

For a = 3 and b = 2: side 5, areas 9 + 6 + 6 + 4 = 25. The drawing is to scale; a and b are positive lengths in the same units.

Algebra repeats this construction:

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

The drawing uses $a=3$ , $b=2$ : $9+6+6+4=25$ , just like $5^2$ . The distributive law proves the identity for any real $a,b$ ; ordinary areas require positive lengths. Deriving squares in OpenStax §1.4.

A squared difference changes the middle sign

Replace $b$ with $-b$ in the formula. The square $(-b)^2=b^2$ , while the mixed product changes sign:

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

Now $99^2=(100-1)^2=10000-200+1=9801$ . Subtracting only $1^2$ gives 9999: that calculates a different expression, $100^2-1^2$ .

A letter may stand for an entire expression: $(2x-3)^2=4x^2-12x+9$ . Here the first square is $(2x)^2$ , and the middle term is $-2\cdot2x\cdot3$ . Parentheses protect both the coefficient and the sign.

A difference of squares cancels mixed terms

When one binomial has a plus and the other a minus, the mixed contributions are opposites:

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

This is a difference of squares, not a squared difference. For example, $103\cdot97=(100+3)(100-3)=10000-9=9991$ . Read it backwards too: $x^2-25=(x-5)(x+5)$ . The product reveals factors hidden in the sum.

Cubes: checking the signs

Multiply $(a-b)$ by $a^2+ab+b^2$ . Six terms appear: $a^3+a^2b+ab^2-a^2b-ab^2-b^3$ . The four middle terms cancel, so:

a^3-b^3=(a-b)(a^2+ab+b^2).

Replacing $b$ with $-b$ gives the sum of cubes:

a^3+b^3=(a+b)(a^2-ab+b^2).

For example, $x^3-8=(x-2)(x^2+2x+4)$ , whereas $x^3+8=(x+2)(x^2-2x+4)$ . The sign on $ab$ makes the middle contributions cancel. Both cube identities, OpenStax §1.5.

Branch · One step furtherAn extra multiplication; the five main identities have already been derived.

Cubing a binomial

Multiply the squared sum by $a+b$ . The term $a^2b$ appears once from $a^2\cdot b$ and twice from $2ab\cdot a$ , giving $3a^2b$ . Similarly, we collect $3ab^2$ :

(a+b)^3=a^3+3a^2b+3ab^2+b^3.

Substituting $-b$ gives $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ . This differs from $a^3-b^3$ : two more contributions lie between the outer cubes.

Special products save writing because every pair of terms has already been counted. If a formula slips your mind, expansion reconstructs it. A small numerical substitution can expose a wrong sign, although it cannot replace a proof.