Exploring Algebraic Identities

Why This Matters

Try a small game. Pick any three “square numbers” that come one after another. The squares of 1, 2, 3 are 1, 4, 9. Now do this: add the smallest and the largest, then take away twice the middle one.

So 1 + 9 = 10. Then 10 − (2 × 4) = 10 − 8 = 2.

Try another set. The squares of 5, 6, 7 are 25, 36, 49. So (25 + 49) − (2 × 36) = 74 − 72 = 2 again!

It is always 2. Every single time. That feels like magic. But it is not magic — it is algebra. By the end of this chapter you will be able to prove why the answer must always be 2, in two short lines.

This chapter is about algebraic identities — a small set of ready-made rules. They do two big jobs for you. They let you square or multiply numbers in your head, like finding 43² without long multiplication. And they let you factorise — break a big expression into smaller multiplied pieces. Both skills show up again and again, in this class and in higher classes too.

The Big Idea

You already know what an equation is. Something like x² − 1 = 24 is an equation. It is only true for some values of x — here, only x = 5 or x = −5. Put in any other number and it breaks.

An identity is special. It is an equation that stays true for every value you put in.

The Big Idea: An identity is an equation that is true for all values of the variables. For example, (a + b)² = a² + 2ab + b² holds no matter what a and b are — positive, negative, fractions, anything. So we can trust it like a tool and use it again and again.

That word “all” is the whole point. Because an identity never fails, you can swap one side for the other any time you like. That swap is what makes calculations and factorising fast.

Seeing (a + b)² as a Picture

Let us start with the most useful identity of all and prove it with a picture.

This identity uses areas of squares and rectangles, so here is a quick reminder of those formulas first.

Imagine two short sticks. One has length a, the other has length b. Join them end to end and you get one stick of length (a + b).

Now build a square whose side is this full length, (a + b). Mark the point where a ends and b begins on each side, and draw lines across. The square breaks neatly into four pieces. Figure 4.1 shows them.

Tap to zoom

Look at what we have. The big square’s area is (a + b)² — that is just side × side. But it is also made of four pieces:

• one square of area a²,
• one square of area b²,
• two rectangles, each of area ab.

Add them: a² + b² + ab + ab = a² + 2ab + b². The pieces must add up to the whole. So we have proved, just by looking, that

(a + b)² = a² + 2ab + b²

This picture only works when a and b are real lengths, so they are positive. But the identity is true for all numbers, even negative ones. Let us check one negative case, then prove it with algebra so we are fully sure.

Check with negatives. Let a = −2 and b = −3. Then (a + b) = −5, so (a + b)² = 25. On the other side: a² = 4, b² = 9, and 2ab = 2 × (−2) × (−3) = 12. Adding, 4 + 12 + 9 = 25. It matches!

The algebra proof (works for every number). We use the distributive property — “multiply each part by each part”:

(a + b)² = (a + b)(a + b)

= a(a + b) + b(a + b)

= a² + ab + ba + b²

= a² + 2ab + b²

This proof never assumed anything about a and b. So it holds for all numbers, and (a + b)² = a² + 2ab + b² is a true identity.

Which is bigger, (a + b)² or a² + b²?

A very common slip is to think (a + b)² is a² + b². It is not. The difference is exactly that extra 2ab.

Take a = 10 and b = 2. Then (a + b)² = 12² = 144, but a² + b² = 100 + 4 = 104. The gap is 40, which is exactly 2ab = 2 × 10 × 2 = 40. When a and b are both positive, 2ab is positive, so (a + b)² is bigger.

Using (a + b)² as a tool

Once we trust the identity, we can use it two ways: to expand expressions, and to square numbers fast.

Here is the identity expanding a binomial — a two-term expression — squared.

Now the number trick. To square a number, split it into an easy round part plus a small part.

Using Identities to Factorise

Factorising means writing an expression as things multiplied together. It is the reverse of expanding. The same identity, read backwards, becomes a factorising tool.

If you ever see an expression shaped like a² + 2ab + b², you can fold it straight back into (a + b)².

Let us factorise x² + 4x + 4. We hunt for the pattern. Notice x² = (x)², and 4 = 2², and the middle term 4x = 2 × (x) × (2). So it matches a² + 2ab + b² with a = x and b = 2. Therefore

x² + 4x + 4 = (x + 2)²

So (x + 2) is a factor of x² + 4x + 4.

Sometimes you first need to pull out a common factor. Here is one of those.

The (a − b)² Identity

What if there is a minus sign? Just replace b with −b in the identity you already have. Since (−b)² = b² and 2a(−b) = −2ab, you get:

(a − b)² = a² − 2ab + b²

Only the middle sign flips. This is also a true identity, and it works the same way for expanding, squaring numbers, and factorising.

We can prove it with a picture too. This time draw a square of side a. Then mark a length b in from one corner, which leaves a length (a − b). The lines cut the square into pieces, shown in Figure 4.2.

Tap to zoom

Reading the figure: the big square has area a². To find the green corner square, which is (a − b)², we subtract the two strips. The bottom strip has area ab. The right strip has area b(a − b). So:

(a − b)² = a² − ab − b(a − b)

= a² − ab − ba + b²

= a² − 2ab + b²

The picture and the algebra agree.

The number trick now uses a minus. It is best when a number is just below a round value.

Proving the “always 2” magic trick

Remember the trick from the start? Now we can prove it in two lines. Any three numbers in a row can be written as (n − 1), n, and (n + 1). Their squares are (n − 1)², n², and (n + 1)².

Add the smallest and largest squares, using our two identities:

(n − 1)² + (n + 1)² = (n² − 2n + 1) + (n² + 2n + 1) = 2n² + 2

Now subtract twice the middle square, which is 2n²:

(2n² + 2) − 2n² = 2

The n² and the −2n and +2n all cancel out, leaving just 2. That is why you always get 2, no matter which three consecutive squares you pick. Magic explained.

The Difference of Two Squares

Here is another identity you met earlier, and it is a real favourite in exams:

(a + b)(a − b) = a² − b²

You can check it by multiplying out: (a + b)(a − b) = a² − ab + ba − b² = a² − b². The middle terms cancel, leaving a² − b².

Written the other way round, a² = (a + b)(a − b) + b², this becomes a clever shortcut for squaring numbers. Figure 4.3 shows why this rearrangement works as areas.

Tap to zoom

A famous Indian mathematician, Śhrīdharāchārya (around 750 CE), used exactly this idea to square numbers fast.

The (a + b + c)² Identity

What if we square the sum of three numbers? We can work it out using a trick: treat b + c as one single block, call it d.

We already know (a + d)² = a² + 2ad + d². Now put d = b + c back in. After expanding 2a(b + c) and (b + c)², and tidying up, we get:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In words: square each of the three terms, then add twice each pair product. There are three pairs — ab, bc, ca — so three “twice” terms.

This has a beautiful picture too. Draw a square of side (a + b + c), and split each side into a, b, and c. You get a 3-by-3 grid of nine pieces, in Figure 4.4.

Tap to zoom

See how the diagonal gives you a², b², c², and each off-diagonal area shows up twice? That is exactly where the 2ab, 2bc and 2ca come from. The picture shows the doubling instead of just stating it.

Multiplying (x + a)(x + b)

Now a different shape. Multiply two brackets that both start with x. Using the distributive property:

(x + a)(x + b) = x² + bx + ax + ab = x² + (a + b)x + ab

So the middle coefficient is the sum a + b, and the last term is the product ab. We can see this with algebra tiles — little squares and strips that stand for x², x, and 1.

Take the example (x + 3)(x + 4). Lay it out as a rectangle with sides (x + 3) and (x + 4), shown in Figure 4.5.

Tap to zoom

The whole rectangle is x² + 7x + 12. Its sides are (x + 3) and (x + 4). Notice the 7 in 7x is split as 3 + 4 — exactly our a + b — and the 12 is 3 × 4 = ab. The tiles make the rule visible.

Factorising x² + (sum)x + (product) by splitting the middle term

To factorise without tiles, just reverse the rule. For x² + bx + c, find two numbers that add to b and multiply to c.

Watch the signs carefully when the numbers are negative.

The Cube Identities

Squares grew into nice patterns. Cubes do too. What is (a + b)³? Use the distributive property, with the square we already know:

(a + b)³ = (a + b)(a + b)² = (a + b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³

This is a brand-new identity. And just as (a + b)² came from cutting a square, (a + b)³ comes from cutting a cube of edge (a + b). Figure 4.6 shows the eight pieces inside.

Tap to zoom

The volume of the whole cube is (a + b)³. Inside, you can count: one cube of volume a³, one of volume b³, three cuboids each a²b, and three each ab². Add them and you get a³ + 3a²b + 3ab² + b³. The volumes prove the identity, exactly the way the areas did before.

For a minus sign, replace b with −b. The signs alternate (plus, minus, plus, minus):

(a − b)³ = a³ − 3a²b + 3ab² − b³

These cube identities let you spot the side of a cube from its volume.

Sum and difference of two cubes

Multiply (x − y)(x² + xy + y²) using the distributive property and almost everything cancels:

(x − y)(x² + xy + y²) = x³ + x²y + xy² − x²y − xy² − y³ = x³ − y³

And similarly:

x³ + y³ = (x + y)(x² − xy + y²)

These two are gold for factorising any difference or sum of cubes. (Notice the sign rule: the middle term of the long bracket is minus for the plus cube identity and plus for the minus one.)

One more identity: a³ + b³ + c³ − 3abc

Multiplying (x + y + z)(x² + y² + z² − xy − yz − zx) and simplifying gives a neat result used in many tricky problems:

x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)

A handy fact falls straight out of it: if x + y + z = 0, then the right side is 0, so x³ + y³ + z³ = 3xyz.

Simplifying Rational Expressions

A rational expression is just one algebraic expression divided by another — a fraction with algebra on top and bottom. The trick to simplify it is the same as with number fractions: factorise top and bottom, then cancel the common factor.

Common Mistakes

Identities are easy to use but also easy to misuse. Here are the slips that trip students most.

Quick Check

Test yourself before the practice set.

Practice Problems

Easy

Medium

Challenge

Summary

You can now explain and use the following:

• An identity is an equation that is true for all values of its variables, unlike an ordinary equation which holds only for special values.
• The square identities (a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b², including their area proofs by cutting a square into pieces.
• The difference of squares (a + b)(a − b) = a² − b², and the three-term square (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca with its grid picture.
• Multiplying (x + a)(x + b) = x² + (a + b)x + ab, and using it backwards to factorise by splitting the middle term.
• The cube identities (a + b)³ and (a − b)³ (with the volume proof from a cube), plus x³ + y³, x³ − y³, and x³ + y³ + z³ − 3xyz.
• Using identities to calculate fast (like 43², 29², 55², 119²) and to simplify rational expressions by factorising and cancelling common factors.

What’s Next

Next we leave algebra for a while and turn to a shape that is everywhere — the circle. You will study its parts, the special angles inside it, and the rules that make wheels, clocks and arches work. Head to Chapter 5 — I’m Up and Down, and Round and Round (Circles).