Algebraic Identities (Extended)

Class 9Polynomials

Algebraic identities are equations that hold true for all values of the variables involved. Unlike equations (which are true for specific values only), identities are universally valid mathematical statements.

In Class 9 Mathematics (NCERT Chapter 2: Polynomials), students extend the basic identities from Class 8 to include cubic identities and three-variable identities. These are more powerful tools for expanding, factorising, and evaluating expressions.

These extended identities are essential for factorisation of polynomials, simplification of complex expressions, and quick evaluation of numerical expressions. They appear frequently in board examinations and competitive tests.

The Class 9 syllabus covers five key extended identities beyond the standard (a + b)² and (a − b)² identities learned in Class 8. Additionally, the important conditional identity involving the sum of three cubes is introduced.

Mastering these identities is critical because they form the foundation for polynomial factorisation, which is heavily used in solving equations, graphing polynomial functions, and simplifying rational expressions in higher classes.

What is Algebraic Identities (Extended)?

Definition: An algebraic identity is a mathematical statement that is true for all values of the variables. Unlike an equation (which is true only for specific values), an identity holds universally.

Class 9 Extended Identities:

Identity I: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Identity II: (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a + b)
Identity III: (a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a − b)
Identity IV: a³ + b³ = (a + b)(a² − ab + b²)
Identity V: a³ − b³ = (a − b)(a² + ab + b²)

Additional identity:

a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
If a + b + c = 0, then a³ + b³ + c³ = 3abc

Algebraic Identities (Extended) Formula

Complete List of Extended Identities:

1. Square of a Trinomial:

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

2. Cube of a Binomial (Sum):

(a + b)³ = a³ + b³ + 3ab(a + b)

3. Cube of a Binomial (Difference):

(a − b)³ = a³ − b³ − 3ab(a − b)

4. Sum of Cubes:

a³ + b³ = (a + b)(a² − ab + b²)

5. Difference of Cubes:

a³ − b³ = (a − b)(a² + ab + b²)

6. Sum of Three Cubes:

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

Derivation and Proof

Derivation of (a + b)³:

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

Derivation of a³ + b³:

From (a + b)³ = a³ + b³ + 3ab(a + b), rearrange:
a³ + b³ = (a + b)³ − 3ab(a + b)
a³ + b³ = (a + b)[(a + b)² − 3ab]
a³ + b³ = (a + b)(a² + 2ab + b² − 3ab)
a³ + b³ = (a + b)(a² − ab + b²)

Derivation of (a + b + c)²:

Let (a + b + c) = [(a + b) + c]
= (a + b)² + 2(a + b)c + c²
= a² + 2ab + b² + 2ac + 2bc + c²
= a² + b² + c² + 2ab + 2bc + 2ca

Derivation of a³ + b³ + c³ − 3abc:

Using a³ + b³ = (a + b)(a² − ab + b²), write a³ + b³ + c³ − 3abc
= (a + b)(a² − ab + b²) + c³ − 3abc
Let s = a + b, so a + b = −c when a + b + c = 0
After algebraic manipulation, the result factors as (a + b + c)(a² + b² + c² − ab − bc − ca)

Types and Properties

Classification of Identities Used in Class 9:

1. Square Identities (from Class 8, used extensively)

(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
a² − b² = (a + b)(a − b)

2. Trinomial Square Identity

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Used when expanding three-variable expressions.

3. Cube Identities

(a + b)³ and (a − b)³ for expanding cubes of binomials.
Used in evaluating expressions like (102)³ or (98)³.

4. Sum and Difference of Cubes

a³ + b³ and a³ − b³ for factorising cubic expressions.
These do NOT reduce to perfect cubes — each has a linear and a quadratic factor.

5. Conditional Identity

If a + b + c = 0, then a³ + b³ + c³ = 3abc.
This is a conditional identity — it holds only when the condition is satisfied.
Very useful for quick evaluation in competitive problems.

Solved Examples

Example 1: Example 1: Expand using (a + b + c)²

Problem: Expand (2x + 3y + 4z)².

Solution:

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

Here a = 2x, b = 3y, c = 4z.

a² = 4x²
b² = 9y²
c² = 16z²
2ab = 2(2x)(3y) = 12xy
2bc = 2(3y)(4z) = 24yz
2ca = 2(4z)(2x) = 16xz

Answer: (2x + 3y + 4z)² = 4x² + 9y² + 16z² + 12xy + 24yz + 16xz

Example 2: Example 2: Find a² + b² + c² given sums

Problem: If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c².

Solution:

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

9² = a² + b² + c² + 2(26)
81 = a² + b² + c² + 52
a² + b² + c² = 81 − 52 = 29

Answer: a² + b² + c² = 29

Example 3: Example 3: Expand (a + b)³

Problem: Expand (2x + 3)³.

Solution:

Using: (a + b)³ = a³ + 3a²b + 3ab² + b³

Here a = 2x, b = 3.

a³ = (2x)³ = 8x³
3a²b = 3(4x²)(3) = 36x²
3ab² = 3(2x)(9) = 54x
b³ = 27

Answer: (2x + 3)³ = 8x³ + 36x² + 54x + 27

Example 4: Example 4: Evaluate using (a − b)³

Problem: Evaluate (98)³ using an identity.

Solution:

Write 98 = 100 − 2. Use (a − b)³ = a³ − 3a²b + 3ab² − b³

Here a = 100, b = 2.

a³ = 1,000,000
3a²b = 3(10000)(2) = 60,000
3ab² = 3(100)(4) = 1,200
b³ = 8
(98)³ = 1,000,000 − 60,000 + 1,200 − 8 = 941,192

Answer: (98)³ = 941,192

Example 5: Example 5: Factorise using a³ + b³

Problem: Factorise 27x³ + 64y³.

Solution:

Write as (3x)³ + (4y)³.

Using: a³ + b³ = (a + b)(a² − ab + b²)

Here a = 3x, b = 4y.

a + b = 3x + 4y
a² = 9x²
ab = 12xy
b² = 16y²

Answer: 27x³ + 64y³ = (3x + 4y)(9x² − 12xy + 16y²)

Example 6: Example 6: Factorise using a³ − b³

Problem: Factorise 125a³ − 8b³.

Solution:

Write as (5a)³ − (2b)³.

Using: a³ − b³ = (a − b)(a² + ab + b²)

Here a = 5a, b = 2b.

a − b = 5a − 2b
a² = 25a²
ab = 10ab
b² = 4b²

Answer: 125a³ − 8b³ = (5a − 2b)(25a² + 10ab + 4b²)

Example 7: Example 7: Evaluate a³ + b³ given sum and product

Problem: If a + b = 7 and ab = 12, find a³ + b³.

Solution:

Using: a³ + b³ = (a + b)³ − 3ab(a + b)

(a + b)³ = 7³ = 343
3ab(a + b) = 3(12)(7) = 252
a³ + b³ = 343 − 252 = 91

Answer: a³ + b³ = 91

Example 8: Example 8: Apply conditional identity a³ + b³ + c³ = 3abc

Problem: If x + y + z = 0, find x³ + y³ + z³ when xy + yz + zx = −11 and xyz = −12.

Solution:

Since x + y + z = 0:

x³ + y³ + z³ = 3xyz
x³ + y³ + z³ = 3(−12) = −36

Answer: x³ + y³ + z³ = −36

Example 9: Example 9: Evaluate (102)³ using identity

Problem: Evaluate (102)³ using (a + b)³.

Solution:

Write 102 = 100 + 2. Use (a + b)³ = a³ + 3a²b + 3ab² + b³

Here a = 100, b = 2.

a³ = 1,000,000
3a²b = 3(10000)(2) = 60,000
3ab² = 3(100)(4) = 1,200
b³ = 8
(102)³ = 1,000,000 + 60,000 + 1,200 + 8 = 1,061,208

Answer: (102)³ = 1,061,208

Example 10: Example 10: Factorise using sum of three cubes

Problem: Factorise x³ + 8y³ + 27z³ − 18xyz.

Solution:

Write as (x)³ + (2y)³ + (3z)³ − 3(x)(2y)(3z).

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

Here a = x, b = 2y, c = 3z.

a + b + c = x + 2y + 3z
a² + b² + c² = x² + 4y² + 9z²
ab + bc + ca = 2xy + 6yz + 3xz

Answer: = (x + 2y + 3z)(x² + 4y² + 9z² − 2xy − 6yz − 3xz)

Real-World Applications

Applications of Extended Algebraic Identities:

Polynomial Factorisation: Identities like a³ + b³ and a³ − b³ are essential for factorising cubic expressions that cannot be handled by simple grouping or splitting the middle term. For example, 8x³ + 27 = (2x)³ + 3³ = (2x + 3)(4x² − 6x + 9).
Quick Numerical Computation: Identities allow rapid calculation of cubes of numbers near round numbers. For example, (997)³ = (1000 − 3)³ can be computed without multiplying 997 three times. This technique saves time in competitive examinations.
Simplification in Mathematical Proofs: Many algebraic, geometric, and number-theoretic proofs require expansion and simplification using these identities. Deriving formulas in coordinate geometry and calculus relies on these building blocks.
Competitive Mathematics (Olympiads, JEE): The conditional identity a³ + b³ + c³ = 3abc (when a + b + c = 0) is a favourite in olympiad-level and JEE problems. It provides elegant shortcuts for problems that would otherwise require lengthy computation.
Physics and Volume Calculations: Cubic relationships appear in volume formulas (V = (4/3)πr³ for a sphere), thermal expansion, and mechanics. Expanding (r + Δr)³ using the cube identity gives the volume change due to a small radius change.
Engineering and Signal Processing: Structural and electrical engineers use polynomial identities when analysing stress distributions, transfer functions, and filter designs. Factorising characteristic polynomials of systems requires these identities.
Financial Mathematics: Compound interest calculations involve cubic and higher-degree polynomials. The identities help in simplifying expressions for three-year compounded growth: (1 + r)³ = 1 + 3r + 3r² + r³.

Key Points to Remember

An identity is true for all values of the variables; an equation is true only for specific values. Example: (a + b)² = a² + 2ab + b² is an identity; 2x + 3 = 7 is an equation.
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca — remember to include all three cross terms. A common mistake is to write only 2ab and forget 2bc and 2ca.
(a + b)³ = a³ + b³ + 3ab(a + b); this compact form is easier to remember than the expanded a³ + 3a²b + 3ab² + b³.
(a − b)³ = a³ − b³ − 3ab(a − b); the signs alternate: +, −, +, − in the expanded form.
a³ + b³ = (a + b)(a² − ab + b²) — the second factor has a minus sign before ab. Do NOT write a² + ab + b² here.
a³ − b³ = (a − b)(a² + ab + b²) — the second factor has a plus sign before ab. Opposite sign convention from sum of cubes.
If a + b + c = 0, then a³ + b³ + c³ = 3abc. This is a conditional identity and is extremely useful for quick evaluation.
To evaluate large cubes, write the number as (round number ± small number) and apply the cube identity. For example: (999)³ = (1000 − 1)³.
Do NOT confuse a³ + b³ with (a + b)³. The difference is: (a + b)³ − (a³ + b³) = 3ab(a + b).
The sign pattern in factorised forms: a³ + b³ has (a + b) as first factor with minus in second; a³ − b³ has (a − b) as first factor with plus in second.
All identities can be verified by expanding the right-hand side or by substituting specific numerical values.
These identities are used in Class 10 (quadratic equations), Class 11 (binomial theorem), and competitive exams (JEE, NTSE, Olympiads).

Practice Problems

Expand (x + 2y + 3z)² using the trinomial identity.
If a + b + c = 12 and a² + b² + c² = 50, find ab + bc + ca.
Expand (3x − 2y)³ using the cube of binomial identity.
Evaluate (999)³ using (a − b)³ identity.
Factorise 64x³ − 125y³.
If x + y = 5 and xy = 6, find x³ + y³.
Verify that 1³ + (−1)³ + 0³ = 3(1)(−1)(0) using the conditional identity.
Factorise 8a³ + 27b³ + 125c³ − 90abc.

Frequently Asked Questions

Q1. What are algebraic identities?

Algebraic identities are equations that are true for all values of the variables. For example, (a + b)² = a² + 2ab + b² holds for every value of a and b, making it an identity.

Q2. What is the difference between an identity and an equation?

An identity is true for all values of the variables (e.g., (a + b)² = a² + 2ab + b²). An equation is true only for specific values (e.g., 2x + 3 = 7 is true only when x = 2).

Q3. How many identities are taught in Class 9?

Class 9 NCERT introduces five extended identities: (a + b + c)², (a + b)³, (a − b)³, a³ + b³, and a³ − b³, along with the conditional identity a³ + b³ + c³ − 3abc.

Q4. What is the expansion of (a + b)³?

(a + b)³ = a³ + 3a²b + 3ab² + b³. This can also be written as a³ + b³ + 3ab(a + b).

Q5. What is the factorisation of a³ + b³?

a³ + b³ = (a + b)(a² − ab + b²). Note the minus sign before ab in the second factor.

Q6. When does a³ + b³ + c³ = 3abc?

This holds when a + b + c = 0. Since a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca), the expression equals 0 when a + b + c = 0.

Q7. How do you use identities to evaluate (101)³?

Write 101 = 100 + 1. Apply (a + b)³ with a = 100, b = 1: (101)³ = 100³ + 3(100²)(1) + 3(100)(1) + 1 = 1,000,000 + 30,000 + 300 + 1 = 1,030,301.

Q8. What is the difference between a³ + b³ and (a + b)³?

a³ + b³ = (a + b)(a² − ab + b²), while (a + b)³ = a³ + b³ + 3ab(a + b). The difference is (a + b)³ − (a³ + b³) = 3ab(a + b).

Q9. Are these identities used in higher classes?

Yes. These identities are fundamental in Class 10 polynomials, Class 11 binomial theorem, and Class 12 calculus. They also appear in competitive exams like JEE and Olympiads.

Q10. How do you verify an identity?

Expand the right-hand side (or left-hand side) and simplify. If both sides are identical for all values of the variables, the equation is an identity. You can also verify by substituting specific values.

Algebraic Identities (Extended)

What is Algebraic Identities (Extended)?

Algebraic Identities (Extended) Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Example 1: Expand using (a + b + c)²

Example 2: Example 2: Find a² + b² + c² given sums

Example 3: Example 3: Expand (a + b)³

Example 4: Example 4: Evaluate using (a − b)³

Example 5: Example 5: Factorise using a³ + b³

Example 6: Example 6: Factorise using a³ − b³

Example 7: Example 7: Evaluate a³ + b³ given sum and product

Example 8: Example 8: Apply conditional identity a³ + b³ + c³ = 3abc

Example 9: Example 9: Evaluate (102)³ using identity

Example 10: Example 10: Factorise using sum of three cubes

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What are algebraic identities?

Q2. What is the difference between an identity and an equation?

Q3. How many identities are taught in Class 9?

Q4. What is the expansion of (a + b)³?

Q5. What is the factorisation of a³ + b³?

Q6. When does a³ + b³ + c³ = 3abc?

Q7. How do you use identities to evaluate (101)³?

Q8. What is the difference between a³ + b³ and (a + b)³?

Q9. Are these identities used in higher classes?

Q10. How do you verify an identity?

Related Topics

More Maths Topics

Quick Links

Algebraic Identities (Extended)

What is Algebraic Identities (Extended)?

Algebraic Identities (Extended) Formula

Derivation and Proof

Types and Properties

Solved Examples

Example 1: Example 1: Expand using (a + b + c)&sup2;

Example 2: Example 2: Find a&sup2; + b&sup2; + c&sup2; given sums

Example 3: Example 3: Expand (a + b)&sup3;

Example 4: Example 4: Evaluate using (a &minus; b)&sup3;

Example 5: Example 5: Factorise using a&sup3; + b&sup3;

Example 6: Example 6: Factorise using a&sup3; &minus; b&sup3;

Example 7: Example 7: Evaluate a&sup3; + b&sup3; given sum and product

Example 8: Example 8: Apply conditional identity a&sup3; + b&sup3; + c&sup3; = 3abc

Example 9: Example 9: Evaluate (102)&sup3; using identity

Example 10: Example 10: Factorise using sum of three cubes

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. What are algebraic identities?

Q2. What is the difference between an identity and an equation?

Q3. How many identities are taught in Class 9?

Q4. What is the expansion of (a + b)&sup3;?

Q5. What is the factorisation of a&sup3; + b&sup3;?

Q6. When does a&sup3; + b&sup3; + c&sup3; = 3abc?

Q7. How do you use identities to evaluate (101)&sup3;?

Q8. What is the difference between a&sup3; + b&sup3; and (a + b)&sup3;?

Q9. Are these identities used in higher classes?

Q10. How do you verify an identity?

Related Topics

More Maths Topics

Quick Links

Example 1: Example 1: Expand using (a + b + c)²

Example 2: Example 2: Find a² + b² + c² given sums

Example 3: Example 3: Expand (a + b)³

Example 4: Example 4: Evaluate using (a − b)³

Example 5: Example 5: Factorise using a³ + b³

Example 6: Example 6: Factorise using a³ − b³

Example 7: Example 7: Evaluate a³ + b³ given sum and product

Example 8: Example 8: Apply conditional identity a³ + b³ + c³ = 3abc

Example 9: Example 9: Evaluate (102)³ using identity

Q4. What is the expansion of (a + b)³?

Q5. What is the factorisation of a³ + b³?

Q6. When does a³ + b³ + c³ = 3abc?

Q7. How do you use identities to evaluate (101)³?

Q8. What is the difference between a³ + b³ and (a + b)³?