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1602 LearnersLast updated on November 13, 2025

The equations used in the algebraic expressions are the algebraic identities. The identities involve operations like addition, multiplication, subtraction, and division. It is used to solve the equations, analyze the data, and calculate area and volume.
The history of algebraic identities begins with ancient civilizations like the Babylonians and Egyptians, who solved equations using words and basic methods such as false position. The Babylonians were especially advanced, handling cubic and quadratic equations using addition and multiplication. Al-Khwarizmi, an Islamic mathematician, is known as the father of algebra and played a key role in developing algebraic equations.
The modern form of algebra, including the use of symbols and variables, was further developed during the Medieval and Renaissance periods by mathematicians like René Descartes and Pierre de Fermat, who introduced modern notation and alphabets as variables.
The LHS (left-hand side) and RHS (right-hand side) of algebraic identities are the same. The identities remain constant even after the value changes. Algebraic identities consist of variables, numbers, and operations. Some standard algebraic identities are:
Let us take a look at the algebraic identities formula of some of the common algebraic identities in this section.
Some of the important algebraic formula identities that we will be using are given as;
1. Square identities
\((a + b)^2 = a^2 + 2ab + b^2 \)
\((a - b)^2 = a^2 - 2ab + b^2 \)
\((a + b)(a - b) = a^2 - b^2 \)
2. Cube identities
\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
\((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)
3. Sum and difference of cubes
\(a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
\(a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
4. Square of trinomial
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
5. Cube of trinomial
\((a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a) \)
\((a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 6abc \)
6. Product of two binomials
\((x + a)(x + b) = x^2 + (a + b)x + ab \)
\((x - a)(x - b) = x^2 - (a + b)x + ab \)
7. Special expansion and factorization
\(a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2)(a^2 - ab + b^2) \)
\(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \)
8. General identity for any number of terms
\((x_1 + x_2 + x_3 + \dots + x_n)^2 = \sum_{i=1}^{n} x_i^2 + 2\sum_{i<j} x_i x_j \)
Algebraic identities make it easier for students to solve algebraic equations. In this section, let’s learn the types of all algebraic identities.
Square of a binomial: The binomial expression has only two terms in it. The square of a binomial is squaring a sum or difference of two terms.
That is, \((a + b)^2 = a^2 + 2ab + b^2\) and \((a - b)^2 = a^2 - 2ab + b^2\)
Difference of squares: The difference of squares is the difference between the perfect squares.
That is, \(a^2 - b^2 = (a + b) (a - b)\)
Cube of a binomial: The cube of a binomial is the addition or subtraction of the third power of the binomial terms.
The formulas are:
\((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
\((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\)
Sum of cubes: The sum of cubes is the sum of two perfect cubes.
They are in the form, \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)
Difference of cubes: The difference between two perfect cubes; which means \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
Perfect square trinomial: The result of squaring a binomial is of three terms, which is a perfect square trinomial.
That is, \(a^2 + 2ab + b^2 = (a + b)^2\)


Proof of algebraic identities helps us in knowing why the identities are true and how it came to be like that. Let use try to prove the algebraic identities.
1. Proof of \(\pmb{(a + b)^2 = a^2 + 2ab + b^2}\)
\((a + b)^2 = (a + b)(a + b) \\[1em] \text{Expand the equation,}\\[1em] (a + b)^2= a(a + b) + b(a + b) \\[1em] (a + b)^2= a^2 + ab + ba + b^2 \\[1em] \text{Since} \ ab=ba,\\[1em] (a + b)^2= a^2 + 2ab + b^2\)
Hence, proved.
2. Proof of \(\pmb{(a - b)^2 = a^2 - 2ab + b^2}\)
\((a - b)^2 = (a - b)(a - b) \\[1em] \text{Expand the equation,}\\[1em] (a - b)^2= a(a - b) - b(a - b) \\[1em] (a - b)^2= a^2 - ab - ba + b^2 \\[1em] \text{Since}\ ab=ba,\\[1em] (a - b)^2= a^2 - 2ab + b^2 \)
Hence, proved.
3. Proof of \(\pmb{(a + b)(a - b) = a^2 - b^2}\)
\((a + b)(a - b) = a(a - b) + b(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a - b) = a^2 - ab + ba - b^2 \\[1em] \text{Upon simplifying},\\[1em] (a + b)(a - b) = a^2 - b^2\)
Hence, proved.
4. Proof of \(\pmb{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}\)
\((a + b)^3 = (a + b)(a + b)(a + b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)^3= (a^2 + 2ab + b^2)(a + b) \\[1em] (a + b)^3= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)^3= a^3 + 3a^2b + 3ab^2 + b^3 \)
Hence, proved.
5. Proof of \(\pmb{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}\)
\( (a - b)^3 = (a - b)(a - b)(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)^3 = (a^2 - 2ab + b^2)(a - b) \\[1em] (a - b)^3 = a^3 - a^2b - 2a^2b + 2ab^2 + ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)
Hence, proved.
6. Proof of \(\pmb{a^3 + b^3 = (a + b)(a^2 - ab + b^2)}\)
\((a + b)(a^2 - ab + b^2) = a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 + b^3 \)
Hence, proved.
7. Proof of \(\pmb{a^3 - b^3 = (a - b)(a^2 + ab + b^2)}\)
\((a - b)(a^2 + ab + b^2) = a(a^2 + ab + b^2) - b(a^2 + ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 - b^3 \)
Hence, proved.
Verifying an algebraic identity simply means that both sides of the equation are equal for all values of the variables. We have some systematic ways to verify an equation, depending on the expression types. Some of the common methods are given as;
1. Expansion method: The expansion method is one of the most common and standard method to verify an algebraic identity. It can be done by following the given steps.
Step 1: Expanding both the sides of the expression completely.
Step 2: Simplifying each side by combining their like terms.
Step 3: If both sides are identical in the end, the identity is verified.
2. Substitution method: We use this method to check whether the given identity is true or not. It can be done by following the given steps.
Step 1: Substitute some numerical values in place of the variables.
Step 2: Simplify LHS and RHS separately.
Step 3: If we get the same result for LHS and RHS, the identity is true for those values.
3. Factorization method: In this method, both the sides are factored and then compared.
Step 1: Factorize both LHS and RHS into simpler algebraic factors
Step 2: If the factorized forms are the same, then the identity is verified.
4. Using algebraic manipulation and simplification: One can either multiply or divide the expression by a non-zero number on both sides, or we can add or subtract any term on both the sides.
To solve complex equations, students use algebraic identities. So let’s learn the importance of algebraic identities for students.
By learning algebraic identities, students can easily solve the complex equation easier. It is important for students to learn and understand their identities. To make the process easier, let’s discuss some tips and tricks.
When learning the algebraic identities to solve the expression, students tend to repeat the same mistakes. In this section, let’s discuss some common mistakes and the ways to avoid them.
Now we learn about algebraic identities, so let's see how we use it in our real world. In the real world, we use it in the fields of engineering, physics, computer science, and so on.
Problem 1. The length of a square garden is x + 4 units. Find the area of the square garden.
The area of the square is x2 + 8x + 16 square units.
The area of a square = s2
Here, s = x + 4 units
Therefore, area of the square = (x + 4)2
Using the identity, (a +b)2 = a2 + 2ab + b2
(x + 4)2 = x2 + 8x + 16
Thus, the area is x2 + 8x + 16 square units
Problem 2. The difference between the squares of two consecutive numbers is 15. Find the numbers?
The numbers are 7 and 8.
Considering the consecutive numbers as x and x + 1
The difference between the squares = (x +1)2 - x2 = 15
Using the identity, a2 - b2 = (a - b) (a + b)
So, (x +1)2 - x2 = (x + 1 - x) (x + 1 + x) = 1(2x + 1)
2x + 1 = 15
2x = 15 - 1
2x = 14
x = 7
Therefore, the numbers are 7 and 8.
Problem 3. Find the volume of a cube-water tank of side x + 2 unit
The volume of the water tank is x3 + 6x2 + 12x + 8 cubic unit.
The volume of the cube = side3
Here, side = (x + 2)3
Using the identity, (a +b)3 = a3 + 3a2b + 3ab2 + b3
(x + 2)3 = x3 + 3x2(2) + 3x(2)2 = 23
x3 + 6x2 + 12x + 8
The volume of the water tank is x3 + 6x2 + 12x + 8 cubic unit.
Problem 4. Expand (a + b)3 where a = 2x and b = 3
The expanded form is 8x3 + 36x2 + 54x + 27.
Using the cube of binomial identity (a +b)3 = a3 + 3a2b + 3ab2 + b3
That is (2x + 3)3 = (2x)3 + 3(2x)2(3) + 3(2x)(3)2 + 33
= 8x3 + 36x2 + 54x + 27
Problem 5. The sum of the cubes of two numbers is 855. If one number is 7, find the other number.
The other number is 8.
Given,
One number is 7
Let’s consider the other number as x
So, x3 + 73 = 855
x3 + 343 = 855
x3 = 855 - 343
x3 = 512
x = ∛512 = 8
So, the other number is 8.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






