Algebraic Identities are algebraic equation which is always true for every value of the variable in them. The algebraic equations that are valid for all values of variables in them are called algebraic identities. It is used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. They contain variable and constant on both the side of polynomial i.e. LHS and RHS. In algebraic identity, LHS must be equal to RHS.
What is Identity?
Consider the equality (x + 1) (x +2) = x2 + 3x + 2. One can evaluate both sides of this equality for some value of a, say x = 5. For x = 5,
- LHS = (x + 1) (x + 2) = (5 + 1) (5 + 2) = 6 × 7 = 42
- RHS = x2 + 3x + 2 = 52 + 3 × 5 + 2 = 25 + 15 + 2 = 42
Thus, the values of the two sides of the equality are equal for a = 5. One can find that for any value of x, LHS = RHS. Such equality, true for every value of the variable in it, is called an identity. Thus, (x + 1) (x + 2) = x2 + 3x + 2 is an identity.
Standard Identities
All standard Algebraic Identities are derived from the Binomial Theorem. There are number of algebraic identities but few are standard that are listed below.
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 + b2 – 2ab
- (a + b)(a – b) = a2 – b2
- (a + b)3=a3 + b3 + 3ab(a + b)
- (a – b)3=a3 – b3 – 3ab(a – b)
- (a + b + c)2=a2 + b2 + c2 + 2ab + 2bc + 2ca
Methods to Solve Identities
- We can verify algebraic identities by substitution method, in which we can put values in variable place and try to make both sides equal. i.e LHS = RHS.
Example:
(a – 2) (a + 2) = a2 – 22
Now we will start putting value in place of a.
starting with a = 1, (-1) x (3) = -3
then we will put a = 2, 0 x 4 = 0
Here we got a = 1 and a = 2 as the value which satisfy the given question.
- Another method is by manipulating identities which are commonly used:
i. (a + b)2 = a2 + b2 + 2ab
ii. (a – b)2 = a2+ b2 – 2ab
iii. (a + b)(a – b) =a2 – b2
iv. (x + a)(x + b) = x2 + (a + b)x + ab
Proof:
i. (a + b)2 = (a + b) (a + b)
= (a + b) (a) + (a + b) (b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
Hence, LHS = RHS.
ii. (a – b)2 = (a – b) (a – b)
= (a – b) (a) + (a – b) (b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
Hence, LHS = RHS.
iii. (a + b) (a – b) = a (a – b) + b (a – b)
= a2 – ab + ab – b2
= a2 – b2
Hence, LHS = RHS.
Applying Identities
Example 1: Solve (2x + 3) (2x – 3) using algebraic identities?
Solution:
By the algebraic identity (a + b)(a – b) = a2 – b2
We can re-write the given expression as
(2x + 3) (2x – 3) = (2x)2 – (3)2 = 4x2– 9
Example 2: Solve (3x + 5)2 using algebraic identities?
Solution:
By algebraic identity
(a + b)2 = a2 + b2 + 2ab
We can re-write the given expression as;
(3x + 5)2 = (3x)2 + 2(3x)5 + 52
(3x + 5)2 = 9x 2 + 30x + 25
Example 3: Find the product of (x + 1)(x + 1) using standard algebraic identities?
Solution:
(x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the standard form I where a = x and b = 1.
We have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x 2 + 2x + 1
Example 4: Expand (3x – 4y)3 using standard algebraic identities?
Solution:
(3x – 4y)3 is of the standard form VII where a = 3x and b = 4y.
We have,
(3x – 4y)3 = (3x)3 – (4y)3 – 3(3x)(4y)(3x – 4y) = 27x 3 – 64y 3 – 108x2y + 144xy 2