Exploring Algebraic Identities: Important Questions and Answers for Class 9

Important questions on Exploring Algebraic Identities for Class 9 are available in this Maths article. These important questions help students revise key identities and solve problems on algebraic expressions with ease. Exploring Algebraic Identities covers fundamental identities along with their applications in expansion, factorisation, and simplification, all aligned with the CBSE syllabus and the NCERT textbook for Class 9. Whether you follow the standard NCERT textbook or the Ganita Manjari, this material supports quick revision and better performance in examinations.

Table of Contents

• Key Concepts on Exploring Algebraic Identities for Class 9
• Important Questions on Exploring Algebraic Identities for Class 9

Key Concepts on Exploring Algebraic Identities for Class 9

An algebraic identity is an equation that holds true for every possible value of its variables. For example:

(x + 3) = 5 is an equation is true only when x = 2.

(a + b)² = a² + 2ab + b² is an identity is true for every value of a and b, always, without exception.

Here are all the standard algebraic identities you must know

Important Questions on Exploring Algebraic Identities for Class 9

Q1. Expand (5p − 3q)².

Solution: Using Identity II: (a − b)² = a² − 2ab + b², where a = 5p and b = 3q:

(5p − 3q)² = (5p)² − 2(5p)(3q) + (3q)²

= 25p² − 30pq + 9q²

Q2. Find the value of (99)² using an algebraic identity.

Solution: Write 99 = 100 − 1.

(99)² = (100 − 1)²

Using Identity II:

= (100)² − 2(100)(1) + (1)²

= 10000 − 200 + 1

= 9801

Q3. If x + 1/x = 5, find x² + 1/x².

Solution: Square both sides:

(x + 1/x)² = 25

⇒ x² + 2·x·(1/x) + 1/x² = 25

⇒ x² + 2 + 1/x² = 25

⇒ x² + 1/x² = 25 − 2 = 23

Q8. Factorise: 25x² − 49y²

Solution: 25x² − 49y² = (5x)² − (7y)²

Using a² − b² = (a + b)(a − b):

= (5x + 7y)(5x − 7y)

Q4. Evaluate 9.8 × 10.2.

Solution: 9.8 × 10.2 = (10 − 0.2)(10 + 0.2)

= (10)² − (0.2)²

= 100 − 0.04

= 99.96

Q5. Using Identity III, find the value of 8.9² − 1.1².

Solution: a² − b² = (a + b)(a − b)

⇒ 8.9² − 1.1² = (8.9 + 1.1)(8.9 − 1.1)

= (10)(7.8)

= 78

Q6. Find the product: (2x + 5)(2x − 3).

Solution: Let a = 2x, b₁ = 5, b₂ = −3:

(2x + 5)(2x − 3) = (2x)² + (5 + (−3))(2x) + (5)(−3)

= 4x² + 2(2x) − 15

= 4x² + 4x − 15

Q7. Expand (2x − y + 3z)².

Solution: Using Identity V with a = 2x, b = −y, c = 3z:

= (2x)² + (−y)² + (3z)² + 2(2x)(−y) + 2(−y)(3z) + 2(3z)(2x)

= 4x² + y² + 9z² − 4xy − 6yz + 12zx

Q8. If a + b + c = 9 and a² + b² + c² = 35, find ab + bc + ca.

Solution: We know: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

(9)² = 35 + 2(ab + bc + ca)

⇒ 81 − 35 = 2(ab + bc + ca)

⇒ 46 = 2(ab + bc + ca)

⇒ ab + bc + ca = 23

Q9. If a + b + c = 0, show that a² + b² + c² = −2(ab + bc + ca).

Solution: From Identity V:

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

Substituting a + b + c = 0:

⇒ 0 = a² + b² + c² + 2(ab + bc + ca)

∴ a² + b² + c² = −2(ab + bc + ca)

Q10. Evaluate (98)³ using a suitable identity.

Solution: 98 = 100 − 2

(98)³ = (100 − 2)³

= (100)³ − 3(100)²(2) + 3(100)(2)² − (2)³

= 1000000 − 60000 + 1200 − 8

= 941192

Q11. If x + y = 12 and xy = 27, find x³ + y³.

Solution: x³ + y³ = (x + y)³ − 3xy(x + y)

= (12)³ − 3(27)(12)

= 1728 − 972

= 756

Q12. If a − b = 3 and ab = 10, find a³ − b³.

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

= (3)³ + 3(10)(3)

= 27 + 90

= 117

Q13. Evaluate 1000³ + 1 without a calculator.

Solution: 1000³ + 1³ = (1000 + 1)(1000² − 1000 × 1 + 1²)

= 1001 × (1000000 − 1000 + 1)

= 1001 × 999001

Q14. If a + b + c = 6 and ab + bc + ca = 11, find a³ + b³ + c³ − 3abc.

Solution: Using Identity X:

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

First, find a² + b² + c²:

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

⇒ 36 = a² + b² + c² + 22

⇒a² + b² + c² = 14

Now:

a² + b² + c² − ab − bc − ca = 14 − 11 = 3

So:

a³ + b³ + c³ − 3abc = 6 × 3 = 18

Q15. If x + y + z = 0, prove that x³ + y³ + z³ = 3xyz.

Solution: From Identity X:

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

Since x + y + z = 0

∴ x³ + y³ + z³ − 3xyz = 0

∴ x³ + y³ + z³ = 3xyz 

Q16. Without actually calculating, find the value of (28)³ + (−15)³ + (−13)³.

Solution: Let a = 28, b = −15, c = −13.

Check: a + b + c = 28 − 15 − 13 = 0

Since a + b + c = 0, we use the special case:

a³ + b³ + c³ = 3abc

= 3 × 28 × (−15) × (−13)

= 3 × 28 × 195

= 3 × 5460

= 16380

Q17. If a + b + c = 9, ab + bc + ca = 26, find a³ + b³ + c³ − 3abc.

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

81 = a² + b² + c² + 52

⇒ a² + b² + c² = 29

a² + b² + c² − ab − bc − ca = 29 − 26 = 3

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

= 9 × 3

= 27