Algebraic identities are equations that remain true for all values of the variables they contain. These identities help in the factorization of polynomials and make it easier to simplify and compute algebraic expressions. They have variables and constants on both sides of the equation (LHS and RHS), and for an algebraic identity to be valid, the LHS must always be equal to the RHS.
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.
Identities List
All standard Algebraic Identities are derived from the Binomial Theorem. There are a 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 Algebraic Identities
- We can verify algebraic identities by substitution method, in which we can put values in variable places 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.
Vedic Shortcuts
Here's another quick shortcut that helps you solve problems faster using algebraic identities, especially in mental maths.
For Example : Find the 982 using this method
Step 1: Pick a nearest base - 98 is close to 100, so use 100 as the base.
Step 2: Find how much it is below the base 98 is 2 less than 100 → difference = –2
Step 3: Square the difference : (−2)2 = 4
Step 4: Subtract the difference from the original number : 98 − 2 = 96
Step 5: Write the answer Take 96 and attach the square of the difference (04) → 9604
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Standard Algebraic Identities Examples
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
Standard Algebraic Identities Practice Problems
Question 1: Solve (4x + 5)2 using algebraic identities.
Question 2: Factorize x2 - 25 using algebraic identities.
Question 3: Expand (2x - 3)3 using algebraic identities.
Question 4: Find the product of (x + 2) (x - 2) using algebraic identities.
Question 5: Expand (a + b + c)2 using algebraic identities.