CLASS 9 • MATHS • CHAPTER 4
Exploring Algebraic Identities
Short, colourful, handwritten-style notes — by EduGrown
1A Curious Pattern to Begin With
Before learning any rule, let us play a small game with square numbers.
✏️ The trick
Pick any three consecutive square numbers. Add the smallest and the largest, then subtract twice the middle one.
- 1, 4, 9 → (1 + 9) – (2 × 4) = 10 – 8 = 2
- 9, 16, 25 → (9 + 25) – (2 × 16) = 34 – 32 = 2
- 25, 36, 49 → (25 + 49) – (2 × 36) = 74 – 72 = 2
💡 What this chapter is about
Patterns like this are not luck — they come from algebraic identities: special rules that help us
expand, factorise and calculate faster.
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2Equation vs Identity
📝 Definition
An algebraic identity is an equation that is true for every value of the variables in it.
An ordinary equation need not be true for all values — only for some.
An ordinary equation need not be true for all values — only for some.
| Statement | True for… | Type |
|---|---|---|
| x² – 1 = 24 | only x = 5 or –5 | Equation |
| (x + y)² = x² + 2xy + y² | every x and every y | Identity |
🔍 Quick test
Put in any numbers you like. If both sides match every single time → it is an identity. If it matches only for special values → it is just an equation.
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3Visualising (a + b)²
Take two lengths a and b and join them to make a longer segment of length (a + b).
Now build a square of side (a + b) and cut it into pieces.
⭐ Reading the picture
- Whole square → area = (a + b)²
- Big inner square → a² • Small square → b²
- Two rectangles → ab + ab = 2ab
✔ Proof for all numbers
The picture works only for lengths (positive numbers). To be sure it works for negatives and fractions too, we use the distributive property:
⭐ Algebraic proof
(a + b)² = (a + b)(a + b) = a(a + b) + b(a + b)
= a² + ab + ba + b² = a² + 2ab + b²
❌ Very important
(a + b)² is NOT a² + b² — the middle term 2ab decides everything:
- If ab is positive → (a + b)² > a² + b²
- If ab is negative → (a + b)² < a² + b²
- If ab = 0 (i.e. a = 0 or b = 0) → both are equal
✏️ Two quick uses
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- Expanding: (5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)² = 25x² + 20xy + 4y²
- Mental maths: 43² = (40 + 3)² = 1600 + 240 + 9 = 1849
4The Identity (a – b)²
Simply replace b by –b in the previous identity — a brand new identity appears!
⭐ The identity
(a – b)² = a² – 2ab + b²
Only the middle sign changes. The first and last terms stay positive.
📝 Reading the picture
Start with a big square of side a and split a into (a – b) and b. Then take the big square and remove the two rectangles:
(a – b)² = a² – ab – b(a – b) = a² – ab – ba + b² = a² – 2ab + b²
✔ Now the Section-1 mystery is solved!
✏️ Why the answer is always 2
Take three consecutive numbers as (n – 1), n, (n + 1). Their squares are (n–1)², n², (n+1)².
(n – 1)² + (n + 1)² = (n² – 2n + 1) + (n² + 2n + 1) = 2n² + 2
Subtract twice the middle square, i.e. 2n²: (2n² + 2) – 2n² = 2 ✔ Always 2!
💡 Mental maths
29² = (30 – 1)² = 900 – 60 + 1 = 841
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5Square of Three Terms (a + b + c)²
What if there are three terms? Clever idea: bundle two of them together.
⭐ The derivation (3 short steps)
- Let d = b + c, so (a + b + c)² becomes (a + d)²
- (a + d)² = a² + 2ad + d² = a² + 2a(b + c) + (b + c)²
- Open the brackets and tidy up:
💡 Easy way to remember
Square each term (3 terms) + twice every pair (3 pairs) = 6 terms in all.
Watch the signs: if a term is negative, its square stays positive, but the products with it flip sign.
Watch the signs: if a term is negative, its square stays positive, but the products with it flip sign.
✏️ Mental maths
119² = (100 + 10 + 9)²
= 100² + 10² + 9² + 2(100)(10) + 2(100)(9) + 2(10)(9)
= 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161
= 100² + 10² + 9² + 2(100)(10) + 2(100)(9) + 2(10)(9)
= 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161
6a² – b² & Śhrīdharāchārya’s Trick
⭐ Difference of squares
a² – b² = (a + b)(a – b)
Rearranged, it becomes a super-fast squaring tool:
a² = (a + b)(a – b) + b²
📜 A gift from 750 CE
The Indian mathematician Śhrīdharāchārya proposed this as a method to square numbers quickly.
55² = (55 + 5)(55 – 5) + 5² = 60 × 50 + 25 = 3025
Choose b so that (a – b) becomes a round number — then the multiplication is almost free!
💡 Try it yourself
35², 65², 85², 105² … do you spot the pattern for numbers ending in 5?
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7Factorisation Using Identities
Identities can be read backwards. Instead of opening brackets, we put them back!
⭐ The 3-step method
- Check the first term — is it a perfect square? Call its root a.
- Check the last term — is it a perfect square? Call its root b.
- Check the middle term — is it exactly 2ab? If yes → answer is (a ± b)².
✏️ Worked examples
- x² + 4x + 4 = x² + 2(x)(2) + 2² = (x + 2)²
- 36x² + 12x + 1 = (6x)² + 2(6x)(1) + 1² = (6x + 1)²
- 16y² – 24y + 9 = (4y)² – 2(4y)(3) + 3² = (4y – 3)²
💡 Golden rule: take out the common factor FIRST
50p² + 60pq + 18q² looks ugly because 50 is not a perfect square. Take 2 out first:
= 2(25p² + 30pq + 9q²) = 2[(5p)² + 2(5p)(3q) + (3q)²] = 2(5p + 3q)²
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8Factorisation Using Algebra Tiles
Algebra tiles turn factorisation into a jigsaw puzzle. Three tiles are used: an x²-tile (big square), an x-tile (long strip) and a unit tile (small square).
📝 How to read it
- The tiles are arranged into a rectangle.
- Its length and breadth are the two factors: (x + 3) and (x + 4).
- Its area is the expression: x² + 7x + 12.
- Notice 7x has been split as 3x + 4x — 3 strips on the right, 4 strips below.
The same idea works when the x² coefficient is not 1:
⭐ General result
(px + a)(qx + b) = pq·x² + (pb + aq)x + ab
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9Splitting the Middle Term
Tiles are fun, but slow. Here is the pen-and-paper method.
⭐ The master identity
x² + (a + b)x + ab = (x + a)(x + b)
So for x² + (middle)x + (last), find two numbers a and b with
a + b = middle coefficient and a × b = last term.
a + b = middle coefficient and a × b = last term.
✏️ Example: x² + 11x + 30
We need a + b = 11 and ab = 30.
- 2 & 15 → product 30 ✔ but sum 17 ✘
- 3 & 10 → product 30 ✔ but sum 13 ✘
- 5 & 6 → product 30 ✔ and sum 11 ✔
🔍 Sign shortcut
Example: x² – 5x + 6 → last is +, middle is – → both negative → –2 and –3 → (x – 2)(x – 3).
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| Last term | Middle term | Signs of a and b |
|---|---|---|
| + | + | both positive |
| + | – | both negative |
| – | + or – | one +, one – (bigger one takes the middle sign) |
10Cube Identities: (a ± b)³
Multiply (a + b) by (a + b)² and a beautiful new identity appears.
⭐ The two cube identities
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a – b)³ = a³ – 3a²b + 3ab² – b³
In the second one the signs go + – + – — they alternate.
A square gave us pieces of a square; a cube gives us pieces of a cube.
📝 Reading the cube picture
A cube of edge (a + b) breaks into 8 pieces:
- 2 cubes → volumes a³ and b³
- 3 cuboids of size a × a × b → total 3a²b
- 3 cuboids of size a × b × b → total 3ab²
✏️ Reading it backwards
p³ + 6p²q + 12pq² + 8q³
= (p)³ + 3(p)²(2q) + 3(p)(2q)² + (2q)³ = (p + 2q)³
So a cube of this volume has side p + 2q units.
= (p)³ + 3(p)²(2q) + 3(p)(2q)² + (2q)³ = (p + 2q)³
So a cube of this volume has side p + 2q units.
💡 Useful rearrangements
a³ + b³ = (a + b)³ – 3ab(a + b)
a³ – b³ = (a – b)³ + 3ab(a – b)
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11Sum & Difference of Cubes
⭐ Two more identities
x³ – y³ = (x – y)(x² + xy + y²)
x³ + y³ = (x + y)(x² – xy + y²)
💡 How to remember
- The small bracket copies the sign of the question.
- In the big bracket, the middle sign is the opposite.
- The big bracket is almost (x ± y)² — but the middle term is xy, not 2xy.
✏️ Family of factors
x – y is a factor of x² – y², of x³ – y³, of x⁴ – y⁴…
In fact (x – y) is a factor of xn – yn for every n.
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In fact (x – y) is a factor of xn – yn for every n.
12The Identity x³ + y³ + z³ – 3xyz
⭐ The three-variable identity
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
✨ The star shortcut
If x + y + z = 0, the whole right side becomes 0, so
x³ + y³ + z³ = 3xyz
This single line solves a huge number of exam questions instantly.
📝 Partner identity you will need with it
x² + y² + z² = (x + y + z)² – 2(xy + yz + zx)
Use it whenever a question gives you the sum and the sum of products.
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13Simplifying Rational Expressions
📝 What is a rational expression?
A fraction whose numerator and denominator are algebraic expressions, e.g. x² – 7x + 125x² + 5x – 100
⭐ The method
- Factorise the numerator.
- Factorise the denominator (take out common factors first).
- Cancel the common factors — but only if they are not zero.
✏️ Worked example
- Top: x² – 7x + 12 = (x – 3)(x – 4)
- Bottom: 5x² + 5x – 100 = 5(x² + x – 20) = 5(x – 4)(x + 5)
- Cancel the common factor (x – 4):
❌ Never do this
You may cancel factors, never single terms.
x + 3x + 5 can NOT be cancelled to 35 — x is a term here, not a factor.
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x + 3x + 5 can NOT be cancelled to 35 — x is a term here, not a factor.
14All Identities at a Glance
| # | Identity | Mainly used for |
|---|---|---|
| 1 | (x + y)² = x² + 2xy + y² | Squaring, perfect-square factors |
| 2 | (x – y)² = x² – 2xy + y² | Squaring numbers just below a round number |
| 3 | (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx | Three-term squares |
| 4 | (x + y)(x – y) = x² – y² | Fast products, Śhrīdharāchārya’s trick |
| 5 | (x + a)(x + b) = x² + (a + b)x + ab | Splitting the middle term |
| 6 | (ax + b)(cx + d) = ac·x² + (ad + bc)x + bd | Quadratics with x² coefficient ≠ 1 |
| 7 | x³ – y³ = (x – y)(x² + xy + y²) | Difference of cubes |
| 8 | x³ + y³ = (x + y)(x² – xy + y²) | Sum of cubes |
| 9 | (x + y)³ = x³ + 3x²y + 3xy² + y³ | Cubing, cube factors |
| 10 | (x – y)³ = x³ – 3x²y + 3xy² – y³ | Cubing with a minus sign |
| 11 | x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – xz – yz) | Three-variable cubes |
15Common Mistakes to Avoid
❌ Don’t do this
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- Writing (a + b)² = a² + b². The 2ab term is never allowed to vanish.
- Writing (a – b)² = a² – b². It is a² – 2ab + b²; the last term is +b², not –b².
- Forgetting the middle sign in cubes. (a – b)³ has signs + – + –.
- In x³ + y³ = (x + y)(x² – xy + y²), writing 2xy instead of xy in the big bracket.
- Trying the perfect-square test before taking out the common factor (e.g. 50p² + 60pq + 18q²).
- Cancelling terms instead of factors in a rational expression.
- Missing a factor: after getting (x – 12)(x + 8) = 0, remember both x = 12 and x = –8 exist — then reject the one that does not fit (like a negative length).
- In (a + b + c)², writing only 3 terms. There must be 6.
16Chapter Summary — Quick Revision
🏆 All of Chapter 4 in 10 lines
- An identity is an equation true for every value of its variables; an ordinary equation is true only for some values.
- Identities can be visualised using squares, rectangles, cubes and algebra tiles — area and volume become algebra.
- (a + b)² = a² + 2ab + b² and (a – b)² = a² – 2ab + b² — only the middle sign differs.
- (a + b + c)² gives 6 terms: three squares plus twice each of the three pairs.
- a² – b² = (a + b)(a – b); rewritten as a² = (a + b)(a – b) + b² it becomes Śhrīdharāchārya’s fast squaring trick (750 CE).
- Identities work both ways — to expand and to factorise. Always remove the common factor first.
- Splitting the middle term: for x² + px + q find two numbers whose sum is p and product is q.
- Cubes: (a + b)³ = a³ + 3a²b + 3ab² + b³, and (a – b)³ has alternating signs.
- x³ ± y³ always factorises; the big bracket carries the opposite middle sign and the term xy (not 2xy).
- x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx); and if x + y + z = 0 then x³ + y³ + z³ = 3xyz.
✨ Happy Learning — @edugrown ✨