Algebraic Formulas for Class 9 – Basic Concepts and Identities

Algebra is just like assigning values to variables and also understanding relationships between values in a given statement. Algebra employs four mathematical operations, such as addition, subtraction, multiplication, and division. 

How Are Maths Identities Helpful for Class 9 Students?

Maths identities for Class 9 usually are used to make complex calculations simpler without using automated electronic calculators. There are many maths identities that Class 9 students can use to lessen their efforts in complex mathematical problems. The 9th class algebra has a huge list of algebraic identities for Class 9 students to make use of. Most of the algebraic identities are used in evaluating polynomial equations and expressions. Algebraic identities for Class 9 maths can also be used to find the product of larger numbers and also their squares. 

Algebraic Identities for Class 9

Algebraic identities for Class 9 is equality that holds good irrespective of the true values chosen for the calculation. It is very useful in many mathematical and scientific computations involving either a very large number or a very small number. Algebraic formulas for Class 9 are generally represented in the form of variables that can take any desired value. 9th class algebra has generalised identities that can be used in many computations. 

Identities of Maths Class 9 with Proofs:

Identity 1:

The square of sum of any two numbers ‘a’ and ‘b’ is given by (a + b)2 = a2 + b2 + 2ab

Proof:

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

Algebraic identities can be proved using an activity method.

Identity 2:

The square of the difference of any two numbers ‘a’ and ‘b’ is given by (a - b)2 = a2 + b2 - 2ab.

Proof:

(a - b)2 = (a - b) (a - b) = a2 - ab - ba + b2 = a2 + b2 - 2ab

Identity 3:

The product of sum and difference of 2 numbers ‘a’ and ‘b’ is given as (a + b) (a - b) = a2 - b2.

Proof:

(a + b) (a - b) = a2 - ab + ab - b2 = a2 - b2

Identity 4:

(x + a) (x + b) = x2 + x (a + b) + ab 

Proof:

(x + a) (x + b) = x2 + xb + ax + ab = x2 + x (a + b) + ab 

Identity 5: 

(x - a) (x + b) = x2 + x (b - a) - ab 

Proof:

(x - a) (x + b) = x2 + xb - ax - ab = x2 + x (b - a) - ab 

Identity 6:

(x - a) (x - b) = x2 - x (a + b) + ab 

Proof:

(x - a) (x - b) = x2 - xb - ax + ab = x2 - x (a + b) + ab 

Identity 7:

The cube of sum of any two numbers is given by (a + b)3 = a3 + b3 + 3ab (a + b)

Proof:

(a + b)3  = (a + b) (a + b)2 

= (a + b) (a2 + b2 + 2ab) 

= a3 + ab2 + 2a2b + ba2 + b3 + 2ab2 

= a3 + b3 + 3a2b + 3ab2 

= a3 + b3 + 3ab (a + b)

Identity 8: 

The cube of difference of any two numbers is given by (a - b)3 = a3 - b3 - 3ab (a - b)

Proof:

(a - b)3 

= (a - b) (a - b)2 

= (a - b) (a2 + b2 - 2ab) 

= a3 + ab2 - 2a2b - ba2 - b3 + 2ab2

= a3 - b3 - 3a2b + 3ab2 

= a3 - b3 - 3ab (a - b)

Identity 9: 

The square of sum of three numbers ‘a’, ‘b’ and ‘c’ is given by 

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

Proof: 

(a + b + c)2

= (a + b + c) (a + b + c) 

= a2 + ab + ac + ba + b2 + bc + ca + cb + c2

= a2 + b2 + c2 + 2ab +2bc + 2ca

Identities of Maths Class 9 Problems

1. Evaluate the square of 99 and 101 using appropriate algebraic identity.

Solution: 

992 = (100 - 1)2 and 1012 = (100 + 1)2

992 can be solved using the algebraic identity (a - b)2 = a2 + b2 - 2ab.

a = 100 and b = 1

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

(100 - 1)2 = 1002 - 2 x 100 x 1 + 12

     = 10000 + 1 - 200

     = 9801

1012 can be solved using the algebraic identity (a + b)2 = a2 + b2 + 2ab.

a = 100 and b = 1

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

(100 + 1)2 = 1002 + 2 x 100 x 1 + 12

      = 10000 + 200 + 1

      = 10201

2. If the sum of two numbers is 12 and the sum of their cubes is 468, find the product of these numbers using algebraic identities.

Solution: 

Let us consider the two numbers as ‘a’ and ‘b’. 

Sum of the numbers = a + b = 12

Sum of their cubes = a3 + b3 = 468

Product of these two numbers = ab =?

The product of two numbers  can be found using the algebraic identity:

(a + b)3 = a3 + b3 + 3ab (a + b)

(12)3 = 468 + 3ab (12)

1728 = 468 + 3ab (12)

36 ab = 1728 - 468

36 ab = 1260

ab = 1260 / 36

ab = 35

The product of two numbers is 35.