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Class 9 Maths Ganita Manjari Algebraic Identities Question Bank with Answers | Exemplar, Olympiad, HOTS & MCQs

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Class 9 Mathematics • Ganita Manjari • Chapter 4

Exploring Algebraic Identities — Advanced Question Bank

A difficult, source-aligned practice resource covering visual identities, factorisation, algebra tiles, quadratic factorisation by splitting the middle term, square and cube identities, difference and sum of cubes, the three-variable cubic identity, rational algebraic expressions and application problems.

30 MCQs15 One-Mark15 Two-Mark 15 Three-Mark10 Four-Mark5 Case Studies 15 Special Exemplar20 Olympiad/HOTS10 High-Level Thinking

Identity Revision Sheet

Squares

(a+b)²=a²+2ab+b²
(a−b)²=a²−2ab+b²

Three terms

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca

Products & factorisation

a²−b²=(a−b)(a+b)
(x+a)(x+b)=x²+(a+b)x+ab

Cubes

(a±b)³=a³±3a²b+3ab²±b³

Sum / difference of cubes

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

Three-cube identity

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

(px+a)(qx+b)=pqx²+(pb+aq)x+ab
If x+y+z=0, then x³+y³+z³=3xyz
Difficulty design: the core bank is already above routine textbook level. Dedicated Exemplar, Olympiad/HOTS and High-Level Thinking sections add non-routine manipulation, proofs, pattern recognition and multi-identity reasoning. These are newly written level-matched problems, not claims of verbatim reproduction from external books.

Visual Learning — HTML Canvas

JavaScript is limited to these drawings. All answer buttons use native HTML <details> and work without JavaScript.

Canvas 1: Geometric model of (a+b)²=a²+2ab+b².
Canvas 2: Algebra-tile model for x²+7x+12=(x+3)(x+4).
Canvas 3: Difference of squares a²−b²=(a−b)(a+b).
Canvas 4: Schematic decomposition behind (a+b)³.

Section A — 30 MCQs

Conceptual, computational and reasoning-based.

1Core

Which statement best distinguishes an identity from an ordinary equation?

  1. A. An identity is true only for positive variables
  2. B. An identity is true for all permissible values of its variables
  3. C. An identity must contain squares
  4. D. An equation is always false for some values
View answer / solution

Answer: B
An algebraic identity is true for all permissible values of its variables.

2Core

For real a and b, (a+b)² − (a²+b²) equals:

  1. A. ab
  2. B. 2ab
  3. C. −2ab
  4. D. a+b
View answer / solution

Answer: B
Using (a+b)²=a²+2ab+b², the difference is 2ab.

3Exemplar-level

If (a+b)² < a²+b², then:

  1. A. ab>0
  2. B. ab<0
  3. C. ab=0
  4. D. a=b
View answer / solution

Answer: B
The difference is 2ab, so it is negative exactly when ab<0.

4Exemplar-level

If (a+b)² = a²+b², then:

  1. A. a=b
  2. B. a=−b
  3. C. ab=0
  4. D. a+b=1
View answer / solution

Answer: C
Equality gives 2ab=0, hence a=0 or b=0.

5Core

The expansion of (5x+2y)² is:

  1. A. 25x²+4y²
  2. B. 25x²+10xy+4y²
  3. C. 25x²+20xy+4y²
  4. D. 5x²+20xy+2y²
View answer / solution

Answer: C
(5x)²+2(5x)(2y)+(2y)²=25x²+20xy+4y².

6Core

Which expression is a perfect square?

  1. A. x²+6x+8
  2. B. x²+6x+9
  3. C. x²+9x+9
  4. D. x²−6x−9
View answer / solution

Answer: B
x²+6x+9=x²+2·x·3+3²=(x+3)².

7Source-based

50p²+60pq+18q² factorises as:

  1. A. 2(5p+3q)²
  2. B. 2(5p−3q)²
  3. C. (10p+6q)²
  4. D. (5p+3q)²
View answer / solution

Answer: A
Take 2 common: 2(25p²+30pq+9q²)=2(5p+3q)².

8Source-based

The value of 79² using (80−1)² is:

  1. A. 6241
  2. B. 6161
  3. C. 6321
  4. D. 6401
View answer / solution

Answer: A
79²=6400−160+1=6241.

9Core

(a+b+c)² contains how many pairwise product terms after combining like terms?

  1. A. 2
  2. B. 3
  3. C. 4
  4. D. 6
View answer / solution

Answer: B
The pairwise terms are 2ab, 2bc and 2ca.

10Source-based

119² equals:

  1. A. 13961
  2. B. 14061
  3. C. 14161
  4. D. 14261
View answer / solution

Answer: C
(100+10+9)²=14161.

11Core

a²−b² factorises as:

  1. A. (a−b)²
  2. B. (a+b)²
  3. C. (a+b)(a−b)
  4. D. (a−b)(a²+ab+b²)
View answer / solution

Answer: C
This is the difference-of-squares identity.

12Source-based

Using a²=(a+b)(a−b)+b², 55² can be computed as:

  1. A. 60×50+25
  2. B. 55×50+25
  3. C. 60×55−25
  4. D. 50×50+25
View answer / solution

Answer: A
Take a=55,b=5:55²=(60)(50)+25=3025.

13Core

(x+3)(x+4) equals:

  1. A. x²+12x+7
  2. B. x²+7x+12
  3. C. x²+x+12
  4. D. x²+7x+7
View answer / solution

Answer: B
Use (x+a)(x+b)=x²+(a+b)x+ab.

14Core

The factors of x²+11x+30 are:

  1. A. (x+3)(x+10)
  2. B. (x+5)(x+6)
  3. C. (x+2)(x+15)
  4. D. (x−5)(x−6)
View answer / solution

Answer: B
We need numbers with sum 11 and product 30: 5 and 6.

15Core

The factors of x²−5x+6 are:

  1. A. (x+2)(x+3)
  2. B. (x−2)(x−3)
  3. C. (x−1)(x−6)
  4. D. (x+1)(x−6)
View answer / solution

Answer: B
−2 and −3 have sum −5 and product 6.

16Source-based

(px+a)(qx+b) has coefficient of x equal to:

  1. A. pq
  2. B. ab
  3. C. pb+aq
  4. D. pa+qb
View answer / solution

Answer: C
Expansion gives pqx²+(pb+aq)x+ab.

17Core

The expansion of (a+b)³ is:

  1. A. a³+b³+3ab
  2. B. a³+3a²b+3ab²+b³
  3. C. a³+3ab²+b³
  4. D. a³−3a²b+3ab²−b³
View answer / solution

Answer: B
This is the cube-of-a-sum identity.

18Core

The expansion of (a−b)³ has signs:

  1. A. all positive
  2. B. all negative
  3. C. + − + −
  4. D. + + − −
View answer / solution

Answer: C
a³−3a²b+3ab²−b³.

19Source-based

p³+6p²q+12pq²+8q³ is:

  1. A. (p+q)³
  2. B. (p+2q)³
  3. C. (2p+q)³
  4. D. (p−2q)³
View answer / solution

Answer: B
Match with a³+3a²b+3ab²+b³ using a=p,b=2q.

20Source-based

8n³−60n²m+150nm²−125m³ equals:

  1. A. (2n−5m)³
  2. B. (2n+5m)³
  3. C. (4n−5m)³
  4. D. (2n−3m)³
View answer / solution

Answer: A
It exactly matches (a−b)³ for a=2n,b=5m.

21Core

x³−y³ factorises as:

  1. A. (x−y)(x²−xy+y²)
  2. B. (x−y)(x²+xy+y²)
  3. C. (x+y)(x²+xy+y²)
  4. D. (x+y)(x²−xy+y²)
View answer / solution

Answer: B
Difference of cubes: x³−y³=(x−y)(x²+xy+y²).

22Core

x³+y³ factorises as:

  1. A. (x+y)(x²−xy+y²)
  2. B. (x+y)(x²+xy+y²)
  3. C. (x−y)(x²−xy+y²)
  4. D. (x−y)(x²+xy+y²)
View answer / solution

Answer: A
Sum of cubes: x³+y³=(x+y)(x²−xy+y²).

23Olympiad

Which expression is always divisible by x−y for integer powers n≥1?

  1. A. xⁿ+yⁿ
  2. B. xⁿ−yⁿ
  3. C. xⁿ+y
  4. D. x+yⁿ
View answer / solution

Answer: B
The chapter explores that x−y divides xⁿ−yⁿ.

24Olympiad

If x+y+z=0, then x³+y³+z³ equals:

  1. A. 0
  2. B. xyz
  3. C. 3xyz
  4. D. −3xyz
View answer / solution

Answer: C
From x³+y³+z³−3xyz=(x+y+z)(…), the RHS is 0.

25Source-based

If x+y+z=10 and x²+y²+z²=38, then xy+yz+zx equals:

  1. A. 19
  2. B. 31
  3. C. 38
  4. D. 62
View answer / solution

Answer: B
100=38+2(xy+yz+zx), so the sum is 31.

26Source-based

Under the same conditions and xyz=25, x³+y³+z³ equals:

  1. A. 75
  2. B. 145
  3. C. 310
  4. D. 455
View answer / solution

Answer: B
Use x³+y³+z³−3xyz=(10)(38−31)=70, hence cube sum=70+75=145.

27Concept

To simplify a rational algebraic expression by cancelling a factor, that factor must:

  1. A. be zero
  2. B. be non-zero for the allowed values
  3. C. be a constant only
  4. D. occur only in the numerator
View answer / solution

Answer: B
A common factor can be cancelled only under the restriction that it is not zero.

28Core

x²+8x+15 factorises as:

  1. A. (x+1)(x+15)
  2. B. (x+3)(x+5)
  3. C. (x−3)(x−5)
  4. D. (x+2)(x+6)
View answer / solution

Answer: B
3+5=8 and 3×5=15.

29Application

A rectangle has area x²+8x+15. Possible side lengths are:

  1. A. x+3 and x+5
  2. B. x+1 and x+15
  3. C. x−3 and x−5
  4. D. x+4 and x+4
View answer / solution

Answer: A
Factor the area as (x+3)(x+5).

30Olympiad

For natural n, n³−n equals:

  1. A. n(n−1)(n+1)
  2. B. n(n−1)²
  3. C. (n−1)(n+1)
  4. D. n(n²+1)
View answer / solution

Answer: A
n³−n=n(n²−1)=n(n−1)(n+1).

Section B — 15 One-Mark Questions

Concise identity recognition, expansion and factorisation.

1Core

Expand (x+7)².

View answer / solution

x²+14x+49.

2Core

Expand (3a−2b)².

View answer / solution

9a²−12ab+4b².

3Core

Factor 25x²+20xy+4y².

View answer / solution

(5x+2y)².

4Core

Factor 16y²−24y+9.

View answer / solution

(4y−3)².

5Core

Find 41² using an identity.

View answer / solution

(40+1)²=1600+80+1=1681.

6Core

Find 97² using an identity.

View answer / solution

(100−3)²=10000−600+9=9409.

7Core

Write the general identity for (x+a)(x+b).

View answer / solution

x²+(a+b)x+ab.

8Core

Factor x²−r².

View answer / solution

(x−r)(x+r).

9Core

Expand (p+q+r)².

View answer / solution

p²+q²+r²+2pq+2qr+2rp.

10Core

Write x³−8 in factorised form.

View answer / solution

(x−2)(x²+2x+4).

11Core

Write 27a³+b³ in factorised form.

View answer / solution

(3a+b)(9a²−3ab+b²).

12Concept

If x+y+z=0, simplify x³+y³+z³−3xyz.

View answer / solution

0.

13Source-based

Factor x²−x−42.

View answer / solution

(x−7)(x+6).

14Exemplar-level

Find the coefficient of x in (2x+3)(5x−4).

View answer / solution

15x coefficient: 2x(−4)+3(5x)=−8x+15x=7x, so coefficient is 7.

15Concept

State the restriction when cancelling x−4 from a rational expression.

View answer / solution

x−4≠0, i.e. x≠4.

Section C — 15 Two-Mark Questions

Short non-routine applications and identity selection.

1Source-based

Show that (n−1)²+(n+1)²−2n²=2.

View answer / solution

Expand: n²−2n+1+n²+2n+1−2n²=2.

2Exemplar-level

Determine when (a+b)²>a²+b².

View answer / solution

Difference=2ab. Therefore (a+b)²>a²+b² exactly when ab>0.

3Core

Factor 9x²+24xy+16y².

View answer / solution

=(3x)²+2(3x)(4y)+(4y)²=(3x+4y)².

4Core

Factor 49x²+28xy+4y².

View answer / solution

=(7x+2y)².

5Source-based

Evaluate 193² using an identity.

View answer / solution

(200−7)²=40000−2800+49=37249.

6Exemplar-level

Evaluate 104×96 without direct multiplication.

View answer / solution

(100+4)(100−4)=10000−16=9984.

7Core

Factor x²+13x+42.

View answer / solution

Find numbers 6,7. Thus (x+6)(x+7).

8Source-based

Factor x²−r−42? Interpret as r²−r−42 and factor it.

View answer / solution

r²−r−42=(r−7)(r+6).

9Core

Expand (p+3q+7r)².

View answer / solution

p²+9q²+49r²+6pq+42qr+14pr.

10Core

Factor 16s²+25t²−40st.

View answer / solution

(4s−5t)².

11Core

Factor 8x³+27.

View answer / solution

(2x+3)(4x²−6x+9).

12Concept

Simplify (x²−9)/(x−3), x≠3.

View answer / solution

(x−3)(x+3)/(x−3)=x+3.

13Exemplar-level

If x+y=8 and xy=12, find x²+y².

View answer / solution

x²+y²=(x+y)²−2xy=64−24=40.

14Exemplar-level

If a−b=5 and ab=6, find a²+b².

View answer / solution

(a−b)²=a²+b²−2ab, so 25=a²+b²−12; hence 37.

15HOTS

Find the missing term so that 4x²+___+9y² is a perfect square.

View answer / solution

Middle term must be ±2(2x)(3y)=±12xy. For (2x+3y)² use +12xy; for (2x−3y)² use −12xy.

Section D — 15 Three-Mark Questions

Multi-step factorisation, proof and rational-expression work.

1Source-extension

Factor 3a²+4ab+(4/3)b² completely.

View answer / solution

Take 1/3 common: (1/3)(9a²+12ab+4b²)=(1/3)(3a+2b)².

2Source-extension

Factor (9/5)s²+6sv+5v².

View answer / solution

Take 1/5 common: (1/5)(9s²+30sv+25v²)=(1/5)(3s+5v)².

3Source-based

Expand (3x−2y+4z)².

View answer / solution

9x²+4y²+16z²−12xy−16yz+24xz.

4Exemplar-level

Verify whether (a+b−c)²+(a−b+c)²+(a−b−c)²=2a²+2b²+2c² is an identity.

View answer / solution

Expand LHS. It becomes 3a²+3b²+3c²−2ab−2ac−2bc, not generally 2a²+2b²+2c². Hence it is not an identity.

5Source-based

Find 147³ using (150−3)³.

View answer / solution

150³−3(150²)(3)+3(150)(9)−27=3,375,000−202,500+4,050−27=3,176,523.

6Exemplar-level

Factor 27u³−(1/8)v³.

View answer / solution

=(3u)³−(v/2)³=(3u−v/2)(9u²+(3/2)uv+v²/4).

7Source-based

Simplify (x²−7x+12)/(5x²+5x−100), assuming denominator non-zero.

View answer / solution

Numerator=(x−3)(x−4); denominator=5(x−4)(x+5). Cancel x−4 (x≠4): (x−3)/[5(x+5)]. Also x≠−5.

8Source-based

A rectangle has area 2x²+7x+3 and width 2x+1. Find its length.

View answer / solution

2x²+7x+3=(2x+1)(x+3), so length=x+3.

9Source-extension

If x+y+z=5 and xy+yz+zx=10, find x²+y²+z².

View answer / solution

25=x²+y²+z²+20, so x²+y²+z²=5.

10Source-based

Using the previous data, find x³+y³+z³−3xyz.

View answer / solution

Identity gives (x+y+z)[x²+y²+z²−xy−yz−zx]=5(5−10)=−25.

11Source-Olympiad

Show n³−n is divisible by 6 for every natural n.

View answer / solution

n³−n=n(n−1)(n+1), product of three consecutive integers. One is divisible by 3 and at least one is even, so product is divisible by 6.

12Exemplar-level

If x+1/x=3, find x²+1/x².

View answer / solution

Square: x²+2+1/x²=9, so x²+1/x²=7.

13Exemplar-level

If x−1/x=4, find x²+1/x².

View answer / solution

Square: x²−2+1/x²=16, so x²+1/x²=18.

14Source-based

Factor 9a²+b²+4c²−6ab+12ac−4bc.

View answer / solution

Recognise (3a−b+2c)²=9a²+b²+4c²−6ab+12ac−4bc.

15Olympiad

If x+y=6 and x³+y³=72, find xy.

View answer / solution

x³+y³=(x+y)³−3xy(x+y):72=216−18xy, so xy=8.

Section E — 10 Four-Mark Questions

Competency, proof, applications and higher-order algebra.

1Source-Olympiad

If both x−2 and x−1/2 are factors of px²+5x+r, show p=r.

View answer / solution

Let P(x)=px²+5x+r. P(2)=4p+10+r=0 ⇒4p+r=−10. P(1/2)=p/4+5/2+r=0 ⇒p+4r=−10. Subtract:3p−3r=0 ⇒p=r.

2Source-based

A rectangular pool has breadth 4 m less than its length and area 96 m². Find its dimensions by factorisation.

View answer / solution

Let length=x, breadth=x−4. x(x−4)=96 ⇒x²−4x−96=0=(x−12)(x+8). Thus x=12 or −8; reject negative. Length=12 m, breadth=8 m.

3Source-extension

Simplify [(x²+x−6)(x²−7x+12)] / [(x²−6x+8)(x²−9)], stating restrictions.

View answer / solution

Factor: (x+3)(x−2)(x−3)(x−4) / [(x−2)(x−4)(x−3)(x+3)]. All factors cancel, giving 1, provided x≠2,4,3,−3.

4Source-based

If x+y+z=10, xyz=25 and x²+y²+z²=38, find x³+y³+z³.

View answer / solution

First 100=38+2(xy+yz+zx), so xy+yz+zx=31. Then x³+y³+z³−75=10(38−31)=70. Hence cube sum=145.

5Source-based

Factor 64u²+121v²+4w²−176uv−32uw+44vw.

View answer / solution

Compare with (8u−11v−2w)²:64u²+121v²+4w²−176uv−32uw+44vw. Therefore factor is (8u−11v−2w)².

6Exemplar-level

Show that (a+b)³+(a−b)³=2a(a²+3b²).

View answer / solution

Expand both cubes. Odd powers of b cancel: [a³+3a²b+3ab²+b³]+[a³−3a²b+3ab²−b³]=2a³+6ab²=2a(a²+3b²).

7Source-extension

Prove x⁴−y⁴ is divisible by x−y and factor it completely over the usual identities.

View answer / solution

x⁴−y⁴=(x²−y²)(x²+y²)=(x−y)(x+y)(x²+y²). Hence x−y is a factor.

8Olympiad

If a+b+c=0, prove a³+b³+c³=3abc, and use it for a=2,b=3,c=−5.

View answer / solution

From a³+b³+c³−3abc=(a+b+c)(…), RHS=0, so equality follows. For 2,3,−5:8+27−125=−90 and 3·2·3·(−5)=−90.

9Source-application

A square playground has side 40 m. A uniform path of width s m is constructed outside it. Find and factor the area of the path.

View answer / solution

Outer side=40+2s. Path area=(40+2s)²−40²=[(40+2s)−40][(40+2s)+40]=(2s)(80+2s)=4s(s+40) m².

10Source-based

If a number plus its reciprocal is 10/3, find the number(s).

View answer / solution

Let x+1/x=10/3. Multiply by 3x:3x²−10x+3=0=(3x−1)(x−3). Hence x=1/3 or 3.

Section F — 5 Case Studies

Algebra tiles, geometry, cube decomposition and applied factorisation.

Case 1Source-based Visual

Case Study 1 — Algebra Tiles and a Rectangle

A rectangle is formed using one x²-tile, seven x-tiles and twelve unit tiles. The arrangement can be made as a complete rectangle.

  1. a) Write its total area.
  2. b) Split 7x in a way that forms the rectangle.
  3. c) Find the side lengths.
  4. d) State the corresponding identity.
View case-study solutions

a) Write its total area. x²+7x+12.

b) Split 7x in a way that forms the rectangle. 3x+4x.

c) Find the side lengths. x+3 and x+4.

d) State the corresponding identity. x²+7x+12=(x+3)(x+4).

Case 2Source-based Application

Case Study 2 — Saira's Algebra-Tile Rectangle

Saira uses one square of area x², eight x-by-1 strips and fifteen unit squares to form a rectangle.

  1. a) Write the total area.
  2. b) Factor the area.
  3. c) Give possible dimensions.
  4. d) Why do 3 and 5 appear?
View case-study solutions

a) Write the total area. x²+8x+15.

b) Factor the area. (x+3)(x+5).

c) Give possible dimensions. x+3 and x+5.

d) Why do 3 and 5 appear? Their sum is 8 and product is 15.

Case 3Source-based Geometry

Case Study 3 — Cube Decomposition

A cube has edge a+b. It is decomposed into two cubes and six cuboids.

  1. a) What is the volume of the large cube?
  2. b) What are the two cube volumes?
  3. c) What is the total volume of the six cuboids?
  4. d) Hence state the identity.
View case-study solutions

a) What is the volume of the large cube? (a+b)³.

b) What are the two cube volumes? a³ and b³.

c) What is the total volume of the six cuboids? 3a²b+3ab².

d) Hence state the identity. (a+b)³=a³+3a²b+3ab²+b³.

Case 4Competency

Case Study 4 — Pool Design

A rectangular pool has area 96 m² and its breadth is 4 m less than its length.

  1. a) If length is x, write the breadth.
  2. b) Form the equation.
  3. c) Factor the quadratic.
  4. d) State the physical dimensions.
View case-study solutions

a) If length is x, write the breadth. x−4.

b) Form the equation. x(x−4)=96, or x²−4x−96=0.

c) Factor the quadratic. (x−12)(x+8)=0.

d) State the physical dimensions. 12 m by 8 m.

Case 5Special Olympiad

Case Study 5 — Three-Number Identity

Three numbers have sum 10, product 25 and sum of squares 38.

  1. a) Find xy+yz+zx.
  2. b) Find x²+y²+z²−xy−yz−zx.
  3. c) Find x³+y³+z³−3xyz.
  4. d) Find x³+y³+z³.
View case-study solutions

a) Find xy+yz+zx. 31.

b) Find x²+y²+z²−xy−yz−zx. 7.

c) Find x³+y³+z³−3xyz. 10×7=70.

d) Find x³+y³+z³. 70+75=145.

Special Exemplar Challenge — 15 Questions

Non-routine identity selection, hidden perfect squares/cubes, exact simplification and deeper factorisation.

1Special Exemplar

Factor 4x²+12xy+9y²−25z² completely.

View worked solution

First three terms=(2x+3y)². Difference of squares gives (2x+3y−5z)(2x+3y+5z).

2Special Exemplar

If a+b=7 and a²+b²=29, find ab and a³+b³.

View worked solution

49=29+2ab⇒ab=10. Then a³+b³=(a+b)³−3ab(a+b)=343−210=133.

3Special Exemplar

Simplify [(a+b)²−(a−b)²]/(4ab), where ab≠0.

View worked solution

Numerator=4ab, so value=1.

4Special Exemplar

Factor x⁴−16.

View worked solution

x⁴−16=(x²−4)(x²+4)=(x−2)(x+2)(x²+4).

5Special Exemplar

If p−q=4 and pq=5, find p³−q³.

View worked solution

p³−q³=(p−q)³+3pq(p−q)=64+60=124.

6Special Exemplar

Find 1003² without long multiplication.

View worked solution

(1000+3)²=1,000,000+6000+9=1,006,009.

7Special Exemplar

Find 997×1003 using an identity.

View worked solution

(1000−3)(1000+3)=1,000,000−9=999,991.

8Special Exemplar

Factor 12x²−7x−10.

View worked solution

Product 12·(−10)=−120; split −7 as 8−15:12x²+8x−15x−10=4x(3x+2)−5(3x+2)=(4x−5)(3x+2).

9Special Exemplar

If x+1/x=5, find x³+1/x³.

View worked solution

Cube identity: (x+1/x)³=x³+1/x³+3(x+1/x). Hence 125=S+15, so S=110.

10Special Exemplar

If x−1/x=3, find x³−1/x³.

View worked solution

(x−1/x)³=x³−1/x³−3(x−1/x). Thus 27=S−9, so S=36.

11Special Exemplar

Factor a³+b³+c³−3abc when a+b+c is known as a factor.

View worked solution

=(a+b+c)(a²+b²+c²−ab−bc−ca).

12Special Exemplar

If a+b+c=6 and ab+bc+ca=11, find a²+b²+c².

View worked solution

36=a²+b²+c²+22, so sum of squares=14.

13Special Exemplar

If the same numbers also have abc=6, find a³+b³+c³.

View worked solution

Difference term: a²+b²+c²−ab−bc−ca=14−11=3. Thus cube sum−18=6·3=18, so cube sum=36.

14Special Exemplar

Simplify (x³−8)/(x²+2x+4), provided denominator≠0.

View worked solution

x³−8=(x−2)(x²+2x+4), so result=x−2.

15Special Exemplar

Show that 999²−1 is divisible by 1000.

View worked solution

999²−1=(999−1)(999+1)=998×1000, hence divisible by 1000.

Extra Olympiad / HOTS Challenge — 20 Questions

Advanced pattern reasoning, proof, symmetric expressions and identity-based number theory.

1Special Olympiad / HOTS

If x+y=1 and x²+y²=5, find x³+y³.

View worked solution

2xy=1−5=−4⇒xy=−2. Cube sum=(x+y)³−3xy(x+y)=1+6=7.

2Special Olympiad / HOTS

If x+y+z=0 and x²+y²+z²=18, find xy+yz+zx.

View worked solution

0=18+2S⇒S=−9.

3Special Olympiad / HOTS

Under the previous conditions, if xyz=4, find x³+y³+z³.

View worked solution

Since x+y+z=0, cube sum=3xyz=12.

4Special Olympiad / HOTS

Prove that x⁵−y⁵ is divisible by x−y and find the quotient.

View worked solution

x⁵−y⁵=(x−y)(x⁴+x³y+x²y²+xy³+y⁴), verified by multiplication.

5Special Olympiad / HOTS

If a+b=10 and ab=21, compute a⁴+b⁴ without finding a,b.

View worked solution

a²+b²=100−42=58. Then a⁴+b⁴=(a²+b²)²−2a²b²=58²−2·441=3364−882=2482.

6Special Olympiad / HOTS

If p+q=4 and pq=1, find p⁵+q⁵.

View worked solution

Let Sₙ=pⁿ+qⁿ. S₀=2,S₁=4 and Sₙ=4Sₙ₋₁−Sₙ₋₂. S₂=14,S₃=52,S₄=194,S₅=724.

7Special Olympiad / HOTS

Find all real x such that (x+3)²+(x−3)²=2x²+18.

View worked solution

Expanding LHS gives x²+6x+9+x²−6x+9=2x²+18, so it is an identity: all real x.

8Special Olympiad / HOTS

Determine k so that x²+kx+36 is a perfect square.

View worked solution

If (x±6)²=x²±12x+36, then k=12 or −12.

9Special Olympiad / HOTS

Determine k so that 4x²+kxy+25y² is a perfect square.

View worked solution

(2x±5y)²=4x²±20xy+25y², so k=±20.

10Special Olympiad / HOTS

If x²+y²=34 and xy=15, find |x−y|.

View worked solution

(x−y)²=34−30=4, so |x−y|=2.

11Special Olympiad / HOTS

If a+b+c=9 and a²+b²+c²=35, find ab+bc+ca.

View worked solution

81=35+2S⇒S=23.

12Special Olympiad / HOTS

If also abc=10, find a³+b³+c³.

View worked solution

Cube difference=(a+b+c)(35−23)=9·12=108. Add 3abc=30, giving 138.

13Special Olympiad / HOTS

Prove n³−n is divisible by 24 when n is odd.

View worked solution

n³−n=n(n−1)(n+1). For odd n, n−1 and n+1 are consecutive even numbers; one is divisible by 4 and the other by 2, giving factor 8. Among three consecutive integers one is divisible by 3. Hence product divisible by 24.

14Special Olympiad / HOTS

Find the remainder of 1001³−1001 when divided by 6.

View worked solution

n³−n is divisible by 6 for every integer n, so remainder 0.

15Special Olympiad / HOTS

Show (a+b)⁴−(a−b)⁴=8ab(a²+b²).

View worked solution

Treat as difference of squares: [(a+b)²−(a−b)²][(a+b)²+(a−b)²]=(4ab)[2a²+2b²]=8ab(a²+b²).

16Special Olympiad / HOTS

If x+1/x=2, prove x=1 without solving a quadratic formula.

View worked solution

x+1/x−2=0⇒(x²−2x+1)/x=0⇒(x−1)²/x=0. Since x≠0, (x−1)²=0⇒x=1.

17Special Olympiad / HOTS

If x+1/x=−2, find x.

View worked solution

Similarly x+1/x+2=0⇒(x+1)²/x=0, so x=−1.

18Special Olympiad / HOTS

Factor x⁶−y⁶ completely using identities.

View worked solution

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

19Special Olympiad / HOTS

If a+b+c=0, prove a²+b²+c²=−2(ab+bc+ca).

View worked solution

Square a+b+c=0: a²+b²+c²+2(ab+bc+ca)=0.

20Special Olympiad / HOTS

Using the previous result, rewrite a³+b³+c³ in terms of abc.

View worked solution

From the three-cube identity and a+b+c=0, a³+b³+c³−3abc=0, hence a³+b³+c³=3abc.

High-Level Thinking Laboratory — 10 Questions

Explain, challenge, generalise and reason beyond mechanical expansion.

1High-Level Thinking

A student claims (a+b)²=a²+b² because 'squaring distributes over addition'. Give a counterexample and then state the exact condition under which the claim becomes true.

View worked solution

Counterexample: a=b=1 gives 4≠2. Exact condition:2ab=0, so a=0 or b=0.

2High-Level Thinking

Can (a+b)² ever be smaller than both a² and b² simultaneously? Give an example or prove impossible.

View worked solution

Yes. Take a=1,b=−1; then (a+b)²=0, which is smaller than both 1 and 1.

3High-Level Thinking

A quadratic x²+mx+n factors as (x+r)(x+s). Explain what information about r,s is encoded in m,n.

View worked solution

m=r+s and n=rs. Thus factorisation is equivalent to finding two numbers with prescribed sum and product.

4High-Level Thinking

Why is cancelling x−4 from a fraction not merely a visual deletion? Explain the domain issue.

View worked solution

Cancellation means dividing numerator and denominator by the same non-zero quantity. At x=4 the factor is zero, division by it is invalid and the original denominator may be zero. Thus x=4 must remain excluded.

5High-Level Thinking

Without expanding fully, decide whether (a+b+c)²+(a−b−c)² is even as a polynomial in a,b,c and simplify it.

View worked solution

Let u=b+c. Then (a+u)²+(a−u)²=2a²+2u²=2[a²+(b+c)²], hence every coefficient is even.

6High-Level Thinking

A number is 3 more than another and their product is 40. Use factorisation to find the numbers.

View worked solution

Let smaller=x, larger=x+3. x(x+3)=40⇒x²+3x−40=(x+8)(x−5)=0. Pairs are (5,8) or (−8,−5).

7High-Level Thinking

For positive a,b with fixed sum S, use an identity to explain when a²+b² is smallest.

View worked solution

a²+b²=(a+b)²−2ab=S²−2ab. For fixed S, this is smallest when ab is largest, which occurs at a=b=S/2. Equivalently (a−b)²≥0.

8High-Level Thinking

Show that (a+b+c)²≥3(ab+bc+ca) for all real a,b,c.

View worked solution

Difference = a²+b²+c²−ab−bc−ca = 1/2[(a−b)²+(b−c)²+(c−a)²]≥0.

9High-Level Thinking

When does equality hold in the previous inequality?

View worked solution

Exactly when a=b=c, because all three squared differences must be zero.

10High-Level Thinking

A cube has volume x³+6x²+12x+8. Without expanding a cube root algorithm, find its edge and explain the pattern.

View worked solution

Recognise x³+3x²·2+3x·2²+2³=(x+2)³. Edge=x+2.

Exam Strategy & Common Traps

TopicBest first moveCommon trap
Perfect-square trinomialCheck first/last squares, then whether middle term is ±2ab.Matching only the first and last terms.
Quadratic factorisationFor x²+Bx+C, find numbers with sum B and product C.Checking product but not sum.
General ax²+bx+cSplit middle term using factors of ac whose sum is b.Using factors of c only.
CubesRecognise first and last cubes before checking middle coefficients.Forgetting coefficient 3.
Three-variable identityCompute xy+yz+zx from (x+y+z)² when needed.Trying to determine x,y,z individually.
Rational expressionsFactor numerator and denominator completely first.Cancelling terms instead of common factors.
ApplicationsTranslate area/volume into a factorable polynomial.Keeping mathematically valid negative dimensions.
WordPress: Paste the complete fragment into one Custom HTML block. No external library is used. If WordPress strips <script>, the notes, all questions and all answer accordions still work; only the four Canvas diagrams will be blank.

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