Algebraic identities are equations where the left-hand side equals the right-hand side for all variable values, simplifying expressions and calculations. Two-variable identities such as (a+b)² and (a-b)² are fundamental, with additional cubic identities like (a−b)³. The document includes examples demonstrating the application of these identities to solve problems.

1. ALGEBRAIC IDENTITIES
DELSI SNEKHA A
23-EDM-05

2. ALGEBRAIC IDENTITIES
An Algebraic identity is basically an equation in
which L.H.S. equals R.H.S. for all values of the
variables. Algebraic identities simplify algebraic
expressions and calculations.

3. APPLICATION OF IDENTITIES
The algebraic identities with two variables are known as two variable
identities. Some basic two-variable algebraic identities are as follows:
(a+b)2
= a2
+b2
+2ab
(a-b)2
= a2
+b2
-2ab
(a2
-b2
) = (a+b)(a-b)

4. APPLICATION OF ALGEBRAIC IDENTITIES
Cubic Identities are,
(a−b)3
=a3
−3a2
b+3ab2
−b3
.
(a+b)3
= a3
+3a2
b+3ab2
+b3
.

5. PROBLEM RELATED TO ALGEBRAIC IDENTITY
Find the value of (3a+4c)2
by using (a+b)2
identity.
SOLUTION:
Comparing (3a+4c)2
, with (a+b)2
we have a=3a, b=4c
Now (a+b)2
= a2
+b2
+2ab
(3a+4c)2
=(3a)2
+2(3a)(4c)+(4c)2
=32
a2
+(2×3×4)(a×c)+42
c2
(3a+4c)2
=9a2
+24ac+16c2

6. PROBLEM RELATED TO ALGEBRAIC IDENTITY
Find the value of (98)3
.
SOLUTION:
(98)3
= (100-2)3
Comparing (a-b)3
, we get a=100,b=2
Now (a−b)3
=a3
−3a2
b+3ab2
−b3
.
(100-2)3
=(100)2
+3(100)2
(2)+3(100)(2)2
−(2)3
=1000000+3(10000)(2)+3(100)(4)-8
=1000000+60000+1200-8
(98)3
=941192

7. THANKYOU