Algebraic identities: squares and difference of squares

Multiply $53 \times 47$ in your head. Long multiplication takes half a minute and invites arithmetic slips — but notice that $53$ and $47$ sit three steps above and below the round number $50$ . That symmetry is not a coincidence you have to spot fresh every time; it is a pattern with a name, an algebraic identity, and once you know it the answer drops out in seconds as $50^2 - 3^2 = 2491$ . This article covers the three standard identities of Class 8: the square of a sum $(a+b)^2$ , the square of a difference $(a-b)^2$ , and the difference of squares $(a+b)(a-b)$ . Each is not a rule to memorise blindly but a direct consequence of the distributive law you already know from expanding two binomials. Once derived and understood, they turn awkward-looking squares and products — $997^2$ , $998 \times 1002$ , $8.9^2$ — into a few lines of easy arithmetic, and they set up factorisation, a core Class 8–9 skill, in the chapter ahead.

The core insight

The three products $(a+b)^2$ , $(a-b)^2$ and $(a+b)(a-b)$ expand, by the distributive law, to $a^2+2ab+b^2$ , $a^2-2ab+b^2$ and $a^2-b^2$ respectively, and recognising which pattern a given multiplication matches turns slow arithmetic into a few quick steps.

The core idea

An identity is an equation true for every value of the letters in it — unlike an ordinary equation, which is only true for particular values and has to be solved. The three identities below are always true, for any $a$ and $b$ whatsoever, which is exactly why they are worth memorising and applying on sight.

1. Square of a sum. Expanding $(a+b)(a+b)$ by the distributive law:

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

where: $a$ and $b$ are the two terms of the bracket, $a^2$ and $b^2$ are their individual squares, and $2ab$ is twice their product — the "middle term" that is easiest to forget.

2. Square of a difference. Expanding $(a-b)(a-b)$ the same way, but with a subtraction:

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

where: the last term is still $+b^2$ , never $-b^2$ , because a negative times a negative is positive; only the middle term changes sign compared to the identity above.

3. Difference of squares. Expanding $(a+b)(a-b)$ , the two middle terms are opposites and cancel out completely rather than combining:

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

where: the result has only two terms, not three — this is the tell-tale sign that a product matches this pattern rather than one of the two square identities above. The figure below shows why $(a+b)^2$ has that extra $2ab$ term: it is a geometric picture, not just algebra. A square of side $(a+b)$ splits into a square of side $a$ , a square of side $b$ , and two identical rectangles of area $ab$ each — hence $2ab$ .

Figure

A large square of side a plus b divided into a smaller a by a square a b by b square and two equal a by b rectangles whose combined area is two a b

The two square identities differ only in the sign of the middle term $2ab$ ; the difference-of-squares identity is structurally different, collapsing to two terms because its middle terms cancel. Spotting which of the three shapes a product matches — same bracket twice with a plus, same bracket twice with a minus, or a plus bracket paired with a minus bracket of the same two terms — is the key skill before applying any identity.

Derivation ADeriving all three identities from the distributive law

The goal is to show that all three standard identities are not separate facts to memorise but a single idea, the distributive law $x(y+z) = xy + xz$ , applied twice to a product of two binomials — so that if you ever forget an identity in an exam, you can rebuild it in under a minute from first principles.

Step 1. Start from the general binomial product, which you already know how to expand: $(a+b)(c+d) = ac + ad + bc + bd$ , obtained by distributing each term of the first bracket over the second.

Step 2. For the square of a sum, set $c=a$ and $d=b$ , so the product becomes $(a+b)(a+b)$ . Substituting into the general expansion: $(a+b)(a+b) = a\cdot a + a\cdot b + b\cdot a + b\cdot b = a^2 + ab + ab + b^2$ . Since $ab = ba$ , the two middle terms are identical and combine: $a^2 + 2ab + b^2$ .

Step 3. For the square of a difference, set $c=a$ and $d=-b$ , so the product is $(a-b)(a-b)$ . The same distribution gives $a^2 - ab - ab + b^2$ , since $(-b)\times(-b) = +b^2$ . The two middle terms are again identical (both $-ab$ ) and combine to give $a^2 - 2ab + b^2$ .

Step 4. For the difference of squares, keep the first bracket as $(a+b)$ but pair it with $(a-b)$ instead, i.e. $c=a$ , $d=-b$ in one bracket only: $(a+b)(a-b) = a\cdot a - a\cdot b + b\cdot a - b\cdot b = a^2 - ab + ab - b^2$ . This time the middle terms are opposites, $-ab$ and $+ab$ , and they cancel to zero, leaving only $a^2 - b^2$ . All three results depend on nothing beyond one application of the distributive law, so they hold for any real numbers $a$ and $b$ — which is exactly what makes them identities rather than equations to be solved.

Worked example

Use the difference-of-squares identity to compute $998 \times 1002$ without long multiplication, and check the result against the earlier example of $997^2$ .

Step 1. Notice that $998$ and $1002$ are equally spaced around the round number $1000$ : $998 = 1000 - 2$ and $1002 = 1000 + 2$ .

Step 2. Match this to the pattern $(a+b)(a-b)$ with $a=1000$ , $b=2$ , so apply the identity $(a+b)(a-b) = a^2 - b^2$ :

998 \times 1002 = (1000-2)(1000+2) = 1000^2 - 2^2

Step 3. Compute each piece:

1000^2 = 1{,}000{,}000, \qquad 2^2 = 4

Step 4. Subtract:

998 \times 1002 = 1{,}000{,}000 - 4 = 999{,}996

The product is $998 \times 1002 = 999{,}996$ . As a check, this should be close to but slightly less than $1000^2 = 1{,}000{,}000$ , and $999{,}996$ fits: the "loss" from $1{,}000{,}000$ is exactly $b^2 = 4$ , however far apart $a$ and $b$ are chosen, showing the same shortcut used earlier for $997^2$ and $53\times 47$ works identically here.

Competitive stretch

The same trick extends to decimals: $8.9^2$ is $(9-0.1)^2 = 9^2 - 2(9)(0.1) + 0.1^2 = 81 - 1.8 + 0.01 = 79.21$ — no long multiplication of decimals needed, only the square-of-a-difference identity with $a=9$ , $b=0.1$ .

Common traps

Forgetting the middle term. Writing $(a+b)^2 = a^2+b^2$ drops the $2ab$ entirely; squaring a sum is not the same as summing the squares.
Sign confusion in $(a-b)^2$ . The last term stays $+b^2$ , never $-b^2$ ; only the middle term flips sign compared to $(a+b)^2$ .
Forcing the wrong identity. The difference-of-squares pattern needs one bracket with a plus and one with a minus of the same two terms; it cannot be applied to $(a+b)(a+b)$ or $(a-b)(a-b)$ .
Choosing an awkward split. The shortcut only saves time if $a$ is round and easy to square, like $50$ , $100$ or $1000$ ; splitting $998$ as $(500+498)$ instead of $(1000-2)$ throws away the advantage.

Ask the tutor: pick any three-digit number close to a multiple of 100 and ask the tutor to walk through the fastest identity-based way to square it.

Why do I need to memorise these identities if I can just multiply the brackets out every time?

You can always fall back on the distributive law, and in fact that is exactly how you should rebuild an identity if you forget it — Step 1 to Step 4 of the derivation above takes under a minute. But in a timed exam, recognising the pattern instantly and writing the three-term or two-term answer directly is much faster than expanding $(a+b)(a+b)$ term by term every single time, especially for numeric shortcuts like $997^2$ or $998\times1002$ where speed matters.

How do I know whether to use the square identity or the difference-of-squares identity on a given multiplication?

Look at the two brackets. If both brackets are literally the same, like $(a+b)(a+b)$ or $(a-b)(a-b)$ , you have a square — use $(a+b)^2$ or $(a-b)^2$ and expect three terms. If the two brackets share the same two terms but one has a plus and the other a minus, like $(a+b)(a-b)$ , you have a difference of squares — expect only two terms, since the middle terms cancel. For a numeric shortcut, check whether the two numbers are equally spaced above and below a round middle value; if so, that middle value is $a$ and the spacing is $b$ .

Does the order of a and b matter, and can b be negative?

The identities hold for any real numbers $a$ and $b$ , including negative ones, because the derivation only used the distributive law, which works for all real numbers. Swapping $a$ and $b$ does not change $(a+b)^2$ or $(a+b)(a-b)$ in the sense that the set of values used is the same, but it does change which number plays which role — always match $a$ to the round, easy-to-square number when using the identities as a mental-math shortcut, since that is what makes the calculation fast.

Algebraic identities: squares and difference of squares

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads