Chapter 9 — Algebraic Identities | Class 9 CBSE Maths 2026-27

Section 9.1 — Core of this Chapter

The 8 Standard Algebraic Identities

An algebraic identity is an equation that holds true for ALL values of the variables. These 8 identities must be memorised and applied fluently — they appear in every CBSE question format.

Identity 1

(a+b)² = a²+2ab+b²

Square of a sum. Expand or evaluate (x+y)² quickly.

Expansion & Evaluation

Identity 2

(a−b)² = a²−2ab+b²

Square of a difference. Note: (a−b)² is always ≥ 0.

Expansion & Min/Max

Identity 3

(a+b)(a−b) = a²−b²

Difference of squares. Reverse: a²−b² = (a+b)(a−b).

Factorisation

Identity 4

(x+a)(x+b) = x²+(a+b)x+ab

Product of two binomials with same first term x.

Quadratic Factorisation

Identity 5

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

Square of a trinomial. 6 terms in expansion.

3-variable Expansion

Identity 6

(a+b)³ = a³+3a²b+3ab²+b³

Cube of a sum. Pascal's triangle: coefficients 1,3,3,1.

Cubic Expansion

Identity 7

(a−b)³ = a³−3a²b+3ab²−b³

Cube of a difference. Alternating signs: +,−,+,−.

Cubic Expansion

Identity 8

a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)

The sum-of-cubes identity. Special case: if a+b+c=0, then a³+b³+c³=3abc.

Special value problems

📐 Geometric Model: (a+b)² = a² + 2ab + b²

The big square of side (a+b) contains: one a² square, one b² square, and TWO rectangles each of area ab. Total = a²+2ab+b² ✓

Section 9.2

Using Identities for Expansion & Evaluation

Example 1

Expand (3x + 5y)² and evaluate (102)². [2M]

a

(3x+5y)² = (3x)²+2(3x)(5y)+(5y)² = 9x²+30xy+25y²

b

(102)² = (100+2)² = 100²+2×100×2+2² = 10000+400+4 = 10404

Example 2

If x + 1/x = 5, find x² + 1/x². [2M]

1

Square both sides: (x+1/x)² = 25
x²+2(x)(1/x)+1/x² = 25
x²+2+1/x² = 25
x²+1/x² = 23

x² + 1/x² = 23

Example 3

Expand (x+y+z)² if x+y+z=10 and xy+yz+zx=27. Find x²+y²+z². [2M]

1

(x+y+z)² = x²+y²+z²+2(xy+yz+zx)

2

100 = x²+y²+z² + 2×27 = x²+y²+z²+54
x²+y²+z² = 46

x² + y² + z² = 46

Section 9.3

Factorisation Using Identities

Identities work in reverse for factorisation — recognising which identity applies is the key skill.

🔑 Factorisation Recognition Guide

Expression pattern	Identity to use	Factored form
a² + 2ab + b²	Identity 1 (reverse)	(a+b)²
a² − 2ab + b²	Identity 2 (reverse)	(a−b)²
a² − b²	Identity 3 (reverse)	(a+b)(a−b)
x² + (a+b)x + ab	Identity 4 (reverse)	(x+a)(x+b)
a³ + b³	Derived from I8	(a+b)(a²−ab+b²)
a³ − b³	Derived from I8	(a−b)(a²+ab+b²)

Example 4

Factorise: (a) 4x²−9 (b) x²+8x+16 (c) 27a³−b³. [3M]

a

4x²−9 = (2x)²−3² = (2x+3)(2x−3) Identity 3

b

x²+8x+16 = x²+2(x)(4)+4² = (x+4)² Identity 1

c

27a³−b³ = (3a)³−b³ = (3a−b)(9a²+3ab+b²)

Section 9.4

Quadratic Factorisation

Factorising ax² + bx + c — the splitting-the-middle-term method and using Identity 4.

📐 Splitting Middle Term: ax² + bx + c

Find p, q such that p×q = a×c and p+q = b

Then rewrite bx = px + qx and group.
Example: 2x² + 7x + 3 → p×q=6, p+q=7 → p=6,q=1 → 2x²+6x+x+3 = 2x(x+3)+1(x+3) = (2x+1)(x+3)

Example 5

Factorise: x²+5x+6. [2M]

1

Need p,q: p×q=6, p+q=5 → p=3,q=2

2

x²+3x+2x+6 = x(x+3)+2(x+3) = (x+2)(x+3)

Example 6

Factorise: 6x²+7x−3. [2M]

1

a×c = 6×(−3)=−18. p+q=7: p=9,q=−2

2

6x²+9x−2x−3 = 3x(2x+3)−1(2x+3) = (3x−1)(2x+3)

Section 9.5

Rational Algebraic Expressions

A rational expression is a fraction with polynomials in numerator and denominator. Simplifying requires factorisation and cancellation of common factors.

Example 7

Simplify: (x²−9)/(x+3). [2M]

1

Numerator: x²−9 = (x+3)(x−3) Identity 3

2

(x+3)(x−3)/(x+3) = x−3 (for x ≠ −3)

(x²−9)/(x+3) = x−3 (for x ≠ −3)

Example 8

Simplify: (x²+5x+6)/(x²+3x+2). [2M]

1

Num: x²+5x+6 = (x+2)(x+3). Denom: x²+3x+2 = (x+1)(x+2)

2

= (x+2)(x+3)/[(x+1)(x+2)] = (x+3)/(x+1) (for x ≠ −2)

(x+3)/(x+1) (for x ≠ −2)

⚠️

Always state restrictions: When you cancel (x+2), state "for x ≠ −2" (since x=−2 makes the denominator 0). CBSE expects this restriction to be written for full marks.

More NCERT Style

Additional Worked Examples

Example 9

If a+b+c=9 and a²+b²+c²=35, find ab+bc+ca. [2M]

1

(a+b+c)² = a²+b²+c²+2(ab+bc+ca)

2

81 = 35 + 2(ab+bc+ca) → ab+bc+ca = 23

ab + bc + ca = 23

Example 10

Evaluate 999³ using identity (a−b)³. [3M]

1

999³ = (1000−1)³

2

= 1000³ − 3×1000²×1 + 3×1000×1 − 1³
= 1,000,000,000 − 3,000,000 + 3,000 − 1
= 997,002,999

999³ = 997,002,999

NCERT Exercise

Exercise 9.1 & 9.2 — Solved

Q1 · 2M

Verify: x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx) for x=1,y=2,z=3.

L

LHS: 1+8+27−3(6)=36−18=18

R

RHS: (6)(1+4+9−2−6−3)=(6)(3)=18 ✓

LHS = RHS = 18 ✓ Identity verified

Q2 · 2M

Evaluate 104 × 96 using identities.

1

104×96 = (100+4)(100−4) = 100²−4² = 10000−16 = 9984

104 × 96 = 9984

Q3 · 3M

If x − 1/x = 3, find x³ − 1/x³.

1

Square: (x−1/x)²=9 → x²−2+1/x²=9 → x²+1/x²=11

2

x³−1/x³ = (x−1/x)(x²+1+1/x²) = 3×(11+1) = 36

x³ − 1/x³ = 36

Smart Study

8 Study Tips for Algebraic Identities

1

Memorise all 8 identities perfectly

Write them out daily until automatic. CBSE will test them without hints. Focus on signs: I2 has minus, I7 alternates +−+−.

2

Split middle term: product × sum

For ax²+bx+c: find p,q where p×q=ac and p+q=b. Then split bx=px+qx and group. Works for ALL quadratics.

3

(a+b+c)² has 6 terms

3 squared terms (a²,b²,c²) + 3 cross terms (2ab,2bc,2ca). Symmetry check: all 3 pairs covered by 2× each.

4

x+1/x → x²+1/x²: square and subtract 2

If x+1/x=k, then x²+1/x²=k²−2. If x−1/x=k, then x²+1/x²=k²+2. Very common CBSE pattern.

5

Restrict domains for rational expressions

After cancelling, always state "x ≠ value." CBSE expects this — it's a partial mark.

6

Special case: a+b+c=0 → a³+b³+c³=3abc

Immediate from Identity 8. Very useful for evaluating expressions where a+b+c=0.

7

Geometric verification builds intuition

The area model of (a+b)² shows WHY the identity works. Draw the square and label its 4 parts — great for remembering the formula.

8

Evaluate large numbers cleverly

103² = (100+3)² = 10000+600+9 = 10609. 997×1003 = (1000−3)(1000+3) = 10⁶−9 = 999991. These "trick" evaluations always appear in MCQs.

Quick Reference

Chapter 9 — All 8 Identities at a Glance

#	Identity	Reverse (factorisation)
I1	(a+b)² = a²+2ab+b²	a²+2ab+b² = (a+b)²
I2	(a−b)² = a²−2ab+b²	a²−2ab+b² = (a−b)²
I3	(a+b)(a−b) = a²−b²	a²−b² = (a+b)(a−b)
I4	(x+a)(x+b) = x²+(a+b)x+ab	Split middle term
I5	(a+b+c)² = a²+b²+c²+2ab+2bc+2ca	—
I6	(a+b)³ = a³+3a²b+3ab²+b³	—
I7	(a−b)³ = a³−3a²b+3ab²−b³	—
I8	a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)	a+b+c=0 → a³+b³+c³=3abc

CBSE Pattern Practice

50 Practice Questions

MCQ · 1 Mark · 2 Marks · 3 Marks · 5 Marks · Case-Based — all with full solutions.

Exploring AlgebraicIdentities