1. Chapter at a Glance
Can you work out 103² in your head? Or 98 × 102? With the right algebraic identity it takes seconds: 103² = 100² + 2×100×3 + 3² = 10609, and 98 × 102 = 100² − 2² = 9996. An identity is an equation that is true for every value of its variables. In this chapter we see identities as pictures (areas and volumes), prove them algebraically, and use them to expand, factorise and simplify expressions quickly.
- Difference between an equation and an identity.
- Square identities with area models: (a ± b)², (a + b)(a − b), (x + a)(x + b).
- (a + b + c)² and its use.
- Cube identities with volume models: (a ± b)³, a³ ± b³.
- The identity a³ + b³ + c³ − 3abc and the special case a + b + c = 0.
- Factorisation using identities, and smart mental arithmetic.
2. Equation vs Identity
Equation: 2x + 3 = 11 is true only for x = 4.
Identity: 2(x + 3) = 2x + 6 is true for all x. Both sides are the same expression written differently.
To test a claimed identity, substitute a couple of values. One failure proves it false; but to prove it true for all values we need algebra (or a general geometric argument).
3. Square Identities and Area Models
- I. (a + b)² = a² + 2ab + b²
- II. (a − b)² = a² − 2ab + b²
- III. (a + b)(a − b) = a² − b²
- IV. (x + a)(x + b) = x² + (a + b)x + ab
Picture for III: cut a b × b square from the corner of an a × a square. The L-shaped remainder, area a² − b², can be cut and rearranged into a rectangle of sides (a + b) and (a − b).
Expand (3x − 5y)².
Solution: (3x)² − 2(3x)(5y) + (5y)² = 9x² − 30xy + 25y².
Evaluate (a) 997² (b) 105 × 95 (c) 103 × 107.
Solution: (a) (1000 − 3)² = 1000000 − 6000 + 9 = 994009. (b) 100² − 5² = 9975. (c) (100 + 3)(100 + 7) = 10000 + 1000 + 21 = 11021.
Useful consequences:
- a² + b² = (a + b)² − 2ab = (a − b)² + 2ab
- (a + b)² − (a − b)² = 4ab
- (a + b)² + (a − b)² = 2(a² + b²)
If x + 1/x = 5, find x² + 1/x².
Solution: (x + 1/x)² = x² + 2 + 1/x² → 25 = x² + 1/x² + 2 → x² + 1/x² = 23.
4. Square of a Trinomial
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Area model: a square of side (a + b + c) splits into 9 pieces — 3 squares and 6 rectangles (two each of ab, bc, ca).
Expand (2x − y + 3)².
Solution: 4x² + y² + 9 + 2(2x)(−y) + 2(−y)(3) + 2(3)(2x) = 4x² + y² + 9 − 4xy − 6y + 12x.
5. Cube Identities and Volume Models
- VI. (a + b)³ = a³ + b³ + 3ab(a + b) = a³ + 3a²b + 3ab² + b³
- VII. (a − b)³ = a³ − b³ − 3ab(a − b) = a³ − 3a²b + 3ab² − b³
- VIII. a³ + b³ = (a + b)(a² − ab + b²)
- IX. a³ − b³ = (a − b)(a² + ab + b²)
Volume picture for VI: a cube of edge (a + b) splits into 8 blocks: one a³ cube, one b³ cube, three a × a × b slabs (3a²b) and three a × b × b rods (3ab²). The coefficients 1, 3, 3, 1 also appear in Pascal's triangle.
Expand (2x + 3)³ and evaluate 101³.
Solution: (2x)³ + 3(2x)²(3) + 3(2x)(3)² + 27 = 8x³ + 36x² + 54x + 27. 101³ = 1000000 + 30000 + 300 + 1 = 1030301.
If a + b = 7 and ab = 10, find a³ + b³.
Solution: a³ + b³ = (a + b)³ − 3ab(a + b) = 343 − 210 = 133.
6. The Identity a³ + b³ + c³ − 3abc
a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
Special case: if a + b + c = 0, then a³ + b³ + c³ = 3abc.
Find (−12)³ + 7³ + 5³ without cubing.
Solution: −12 + 7 + 5 = 0, so the sum = 3 × (−12) × 7 × 5 = −1260.
7. Factorisation Using Identities
| Expression | Identity used | Factors |
|---|---|---|
| 9x² − 49 | a² − b² | (3x − 7)(3x + 7) |
| 4x² + 12x + 9 | (a + b)² | (2x + 3)² |
| x² + 7x + 12 | (x + a)(x + b) | (x + 3)(x + 4) |
| 27x³ − 8 | a³ − b³ | (3x − 2)(9x² + 6x + 4) |
| x⁴ − 16 | a² − b² twice | (x − 2)(x + 2)(x² + 4) |
Splitting the middle term for ax² + bx + c: find two numbers whose product is ac and whose sum is b.
Factorise 6x² + 7x − 3.
Solution: ac = −18; numbers 9 and −2 (sum 7). 6x² + 9x − 2x − 3 = 3x(2x + 3) − 1(2x + 3) = (3x − 1)(2x + 3).
8. Solved Problems (Board Pattern)
If x − y = 4 and xy = 21, find x² + y² and x³ − y³.
Solution: x² + y² = 16 + 42 = 58. x³ − y³ = (x − y)³ + 3xy(x − y) = 64 + 252 = 316.
Factorise 8a³ + b³ + 12a²b + 6ab².
Solution: = (2a)³ + 3(2a)²b + 3(2a)b² + b³ = (2a + b)³.
If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².
Solution: 81 = a² + b² + c² + 52 → a² + b² + c² = 29.
9. Master Formula Table
| # | Identity |
|---|---|
| I | (a + b)² = a² + 2ab + b² |
| II | (a − b)² = a² − 2ab + b² |
| III | a² − b² = (a + b)(a − b) |
| IV | (x + a)(x + b) = x² + (a + b)x + ab |
| V | (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca |
| VI | (a + b)³ = a³ + b³ + 3ab(a + b) |
| VII | (a − b)³ = a³ − b³ − 3ab(a − b) |
| VIII | a³ + b³ = (a + b)(a² − ab + b²) |
| IX | a³ − b³ = (a − b)(a² + ab + b²) |
| X | a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) |
10. Master Summary Table
| Type | Model | Exam use |
|---|---|---|
| Square identities | Area of square split into parts | Expand; mental squares |
| Difference of squares | L-shape → rectangle | Products like 98 × 102; factorise |
| Trinomial square | 9-piece square | Find a² + b² + c² |
| Cube identities | 8-piece cube | Find a³ ± b³ from sums/products |
| a³ + b³ + c³ − 3abc | Algebraic | a + b + c = 0 shortcut |
11. Mind Map
- (a ± b)²
- a² − b²
- (x + a)(x + b)
- (a + b + c)²
- (a ± b)³
- a³ ± b³
- Factorise
- Mental maths
12. Tips and Tricks
- Numbers near 100 or 1000: write as 100 ± k and use I, II or III.
- "If a + b and ab are given" → you can find a² + b², a³ + b³, (a − b)².
- Whenever three terms add to zero, think "sum of cubes = 3abc".
- Always take out a common factor first before using an identity.
- (a + b)² ≠ a² + b² — don't forget 2ab.
- Sign errors in (a − b)³: the terms alternate +, −, +, −.
- a³ + b³ has factor (a² − ab + b²), not (a − b)².
13. Important One-liners
- An identity is true for all values of the variables.
- (a + b)² − (a − b)² = 4ab.
- a² − b² = (a + b)(a − b): the most-used factorisation.
- Coefficients of (a + b)³ are 1, 3, 3, 1.
- If a + b + c = 0 then a³ + b³ + c³ = 3abc.
- x² + 1/x² = (x + 1/x)² − 2.
14. Exam-pattern Questions with Detailed Answers
Expand (x + 6)(x − 4).
Show answer
x² + (6 − 4)x + (6)(−4) = x² + 2x − 24.
Answer: x² + 2x − 24
Evaluate 52² − 48² using an identity.
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(52 + 48)(52 − 48) = 100 × 4 = 400.
Answer: 400
Factorise 25a² − 36b².
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(5a)² − (6b)² = (5a − 6b)(5a + 6b).
Answer: (5a − 6b)(5a + 6b)
Evaluate 99³ using an identity.
Show answer
(100 − 1)³ = 1000000 − 30000 + 300 − 1 = 970299.
Answer: 970299
Factorise x² − 11x + 28.
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Two numbers with product 28 and sum −11: −4 and −7. (x − 4)(x − 7).
Answer: (x − 4)(x − 7)
If x − 1/x = 3, find x² + 1/x².
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(x − 1/x)² = x² − 2 + 1/x² → 9 + 2 = 11.
Answer: 11
Factorise 64m³ − 27n³.
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(4m)³ − (3n)³ = (4m − 3n)(16m² + 12mn + 9n²).
Answer: (4m − 3n)(16m² + 12mn + 9n²)
Find (28)³ + (−15)³ + (−13)³ without cubing.
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28 − 15 − 13 = 0, so the sum = 3 × 28 × (−15) × (−13) = 16380.
Answer: 16380
Factorise 2x² + 5x − 12 and hence find its zeros.
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ac = −24; 8 and −3. 2x² + 8x − 3x − 12 = 2x(x + 4) − 3(x + 4) = (2x − 3)(x + 4). Zeros: 3/2 and −4.
Answer: (2x − 3)(x + 4); zeros 3/2, −4
If x + 1/x = 4, find x³ + 1/x³.
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(x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x) → 64 = x³ + 1/x³ + 12 → 52.
Answer: 52
Factorise a³ − b³ + 1 + 3ab.
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Write as a³ + (−b)³ + 1³ − 3(a)(−b)(1). By identity X: (a − b + 1)(a² + b² + 1 + ab + b − a).
Answer: (a − b + 1)(a² + b² + 1 + ab + b − a)
The area of a rectangle is 25a² − 35a + 12. Find possible expressions for its length and breadth.
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ac = 300; −20 and −15 (sum −35). 25a² − 20a − 15a + 12 = 5a(5a − 4) − 3(5a − 4) = (5a − 3)(5a − 4).
Answer: Length 5a − 3, breadth 5a − 4 (or the other way round)
15. Quick Revision Summary
- Identities hold for all values; equations only for some.
- Squares: (a ± b)², a² − b², (x + a)(x + b), (a + b + c)² — see them as area models.
- Cubes: (a ± b)³, a³ ± b³ — see them as volume models.
- a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc.
- Factorise by common factor → identity → splitting the middle term.