Algebraic Identities: Definition, Derivations and Applications

Algebraic Identities: Algebraic equations form the basis of all simple and complex formulas used in Mathematics. The algebraic identities are the equations in which all the values of variables are valid and the equation will hold true irrespective of any value of the variables. Algebraic identities find application in advanced mathematical equations, analysis and research-based concepts. It is important for mathematics students to be proficient with these equations as they are very helpful in solving engineering and scientific problems.

This article covers the four main forms of algebraic identities, their derivation and their applications. Their properties are also important to understanding and grasping various mathematical functions associated with them. The most important applications are found in proving binomial theorems and factorisation of polynomials. The article provides valuable insight on the same. Read the complete article to know more.

Algebraic Identities: Definition

Algebraic identities are algebraic equations that are true for all the values of variables in them. Algebraic identities and expressions are mathematical equations that comprise numbers, variables (unknown values), and mathematical operators (addition, subtraction, multiplication, division, etc.)

Algebraic identities are used in various branches of mathematics, such as algebra, geometry, trigonometry etc. These are mainly used to find the factors of the polynomials. A better understanding of algebraic identities contributes toward strengthening the efficiency to solve problem sums. One of the most important applications of algebraic identities is the factorisation of polynomials.

What are Algebraic Identities?

If an equation is true for all values of the variables in it, it is called an identity. The algebraic identities are the equations in which the value of the left-hand side of an equation identically equals the value of the right-hand side of the equation for all values of the variable.

Example: Consider the linear equation \(ax + b = 0.\)
Here, the left-hand side and right-hand side of the above equations are the same when \(x = – \frac{b}{a}.\) Hence, it is not identity, but it is an equation.
In \({\left( {a + b} \right)^2} = {a^2} + {b^2} + 2\,ab,\) we know that it is true for all values of variables \(a\) and \(b.\) So, it is an identity.

Standard Algebraic Identities

We have some standard identities to use in the various branches of mathematics. All the standard identities are derived by using the Binomial theorem.

Four standard algebraic identities are listed below:

Identity-1: Algebraic Identity of Square of Sum of Two Terms
\({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)

Identity-2: Algebraic Identity of Square of Difference of Two Terms
\({\left( {a – b} \right)^2} = {a^2} – 2\,ab + {b^2}\)

Identity-3: Algebraic Identity of Difference of Two Squares
\(\left( {a + b} \right)\left( {a – b} \right) = {a^2} – {b^2}\)

Identity-4: Algebraic Identity \(\left( {x + a} \right)\left( {x + b} \right)\)
\(\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\)

Algebraic Identities Under Binomial Theorem

The Binomial Theorem hands out a standard way of expanding the powers of binomials or other terms. The binomial theorem is used in algebra, probability, etc. The general form of such algebraic identities is mentioned below:

\({\left( {a + b} \right)^2} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n – 1}}.b + {}^n{C_2}{a^{n – 2}}.{b^2} + …. + {}^n{C_{n – 1}}a.{b^{n – 1}} + {}^n{C_n}{b^n}\)

To find the binomial co-efficient, we will use the pascals triangle:

The identities derived by using binomial theorem are tabulated below:

Factorising Algebraic Identities

We know that algebraic identities are used in factorising polynomials. Some of the below-listed identities are used for the factorisation of algebraic identities and expressions.

1. Identity -1:
\({a^2} – {b^2} = \left( {a + b} \right)\left( {a – b} \right)\)
2. Identity-2:
\({a^3} – {b^3} = \left( {a – b} \right)\left( {{a^2} + ab + {b^2}} \right)\)
3. Identity-3:
\({a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} – ab + {b^2}} \right)\)
4. Identity-4:
\({a^4} – {b^4} = \left( {{a^2} – {b^2}} \right)\left( {{a^2} + {b^2}} \right)\)

Trinomial Algebraic Identities

The corresponding equalities are of trinomial algebraic identities. You can derive such identities simply by factorising and manipulating the terms given below:

1. Identity -1:
\(\left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right) = \left( {a + b + c} \right)\left( {ab + ac + bc} \right) – abc\)
2.Identity-2:
\({a^3} + {b^3} + {c^3} – 3\,abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} – ab – bc – ca} \right)\)
3. Identity-3:
\({a^2} + {b^2} + {c^2} = {\left( {a + b + c} \right)^2} – 2\left( {ab + bc + ca} \right)\)
4. Identity-4:
\(\left( {a – b} \right)\left( {a – c} \right) = {a^2} – \left( {b + c} \right)a + bc\)

Proofs of Basic Algebraic Identities

We can visualise and study the proofs of some of the basic algebraic identities:

Proof of \({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)

Let us consider a square with side \(\left( {a + b} \right)\) units
The big square is divided into four quadrilaterals (rectangles, squares), as shown in the figure.

Using the area of rectangle \(\left( {{\rm{length}} \times {\rm{breadth}}} \right)\) and area of square \({\left( {{\rm{side}}} \right)^2}\) we can visualise the identity as follows:

The total area of the square is the sum of areas of rectangles and squares in it.
\( \Rightarrow {\left( {a + b} \right)^2} = {a^2} + ab + ab + {b^2}\)
\( \Rightarrow {\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)

Proof of \({\left( {a – b} \right)^2} = {a^2} – 2\,ab + {b^2}\)

Let us consider a square with side \(a = \left( {a – b} \right) + b\) units as shown in the figure. The big square is divided into four quadrilaterals (rectangles, squares), as shown in the figure.

Using the area of rectangle \(\left( {{\rm{length}} \times {\rm{breadth}}} \right)\) and area of square \({\left( {{\rm{side}}} \right)^2}\), we can visualise the identity as follows:

The total area of the square is the sum of areas of rectangles and squares in it.

1. \( \Rightarrow {a^2} = {\left( {a – b} \right)^2} + ab + b\left( {a – b} \right)\)
2. \( \Rightarrow {a^2} = {\left( {a – b} \right)^2} + ab + ba – {b^2}\)
3. \( \Rightarrow {a^2} = {\left( {a – b} \right)^2} + 2\,ab – {b^2}\)
4. \( \Rightarrow {\left( {a – b} \right)^2} = {a^2} – 2\,ab + {b^2}\)

Proof of \(\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\)

Let us consider a rectangle with length \(\left( {x + b} \right)\) units and breadth \(\left( {x + a} \right)\) units.
The big rectangle is divided into four quadrilaterals (rectangles, squares), as shown in the figure.

Using the area of rectangle \(\left( {{\rm{length}} \times {\rm{breadth}}} \right)\) and area of square \({\left( {{\rm{side}}} \right)^2}\) we can visualise the identity as follows:

The total area of the rectangle is the sum of areas of rectangles and squares in it.

1. \( \Rightarrow \left( {x + a} \right)\left( {x + b} \right) = {x^2} + ax + bx + ab\)
2. \( \Rightarrow \left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\)

Proof of \({a^2} – {b^2} = \left( {a + b} \right)\left( {a – b} \right)\)

Let us consider a square with side \(a = \left( {a – b} \right) + b\) units as shown in the figure.
The big square is divided into four quadrilaterals (rectangles, squares), as shown in the figure.

Using the area of rectangle \(\left( {{\rm{length}} \times {\rm{breadth}}} \right)\) and area of square \({\left( {{\rm{side}}} \right)^2}\), we can visualise the identity as follows:

The total area of the square is the sum of areas of rectangles and squares in it.

1. \( \Rightarrow {a^2} = a\left( {a – b} \right) + b\left( {a – b} \right) + {b^2}\)
2. \( \Rightarrow {a^2} = \left( {a + b} \right)\left( {a – b} \right) + {b^2}\)
3. \( \Rightarrow {a^2} – {b^2} = \left( {a + b} \right)\left( {a – b} \right)\)

Proof of \({\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\,ab + 2\,bc + 2\,ca\)

Let us consider a square with side \(a + b + c\) units, as shown in the figure.
The big square is divided into nine quadrilaterals (rectangles, squares), as shown in the figure.

Using the area of rectangle \(\left( {{\rm{length}} \times {\rm{breadth}}} \right)\) and area of square \({\left( {{\rm{side}}} \right)^2}\), we can visualise the identity as follows:

The total area of the square is the sum of areas of rectangles and squares in it.

1. \( \Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + ab + bc + ca + ba + cb + ac + {c^2} + {b^2}\)
2. \( \Rightarrow {\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\,ab + 2\,bc + 2\,ca\)

Algebraic Identities Chart

The chart of algebraic identities helps us to understand various types of identities, uses and applications in algebra and other branches of mathematics. The chart includes:
1. Square of Binomial
2. Difference Between Squares
3. Cube of Binomials
4. Sum of Cubes
5. Difference Between Cubes
6. Product of Binomials
7. Square of Trinomials

Solved Example Questions on Algebraic Identities

Q.1. Find the product of \(\left( {x + 2} \right)\left( {x + 2} \right)\) using standard algebraic identities.
Ans:
We can write \(\left( {x + 2} \right)\left( {x + 2} \right)\) as \({\left( {x + 2} \right)^2}.\)
By using the identity: \({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)
By putting the value of \(x = a\) and \(y = 2,\) we get
\( \Rightarrow {\left( {x + 2} \right)^2} = {x^2} + {2^2} + 2 \times x \times 2\)
\( \Rightarrow {\left( {x + 2} \right)^2} = {x^2} + 4 + 4\,x\)
Hence, the value of \(\left( {x + 2} \right)\left( {x + 2} \right)\) is \({x^2} + 4x + 4.\)

Q.2. Factorise \(25\,{x^2} + 16\,{y^2} + 9\,{z^2} – 40\,xy + 24\,yz – 30\,zx\) using standard algebraic identities.
Ans:
Given: \(25\,{x^2} + 16\,{y^2} + 9\,{z^2} – 40\,xy + 24\,yz – 30\,zx\)
\( \Rightarrow {5^2}\,{x^2} + {4^2}\,{y^2} + {3^2}\,{z^2} – 40\,xy + 24\,yz – 30\,zx\)
\( \Rightarrow {\left( { – 5\,x} \right)^2} + {\left( {4\,y} \right)^2} + {\left( {3\,z} \right)^2} + 2\, \times \left( { – 5\,x} \right) \times 4\,y + 2 \times 4\,y \times 3\,z + 2 \times \left( { – 5\,x} \right) \times 3\,z\)
By using the identity: \({\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2\,ab + 2\,bc + 2\,ca\)
By putting the value of \(a = – 5\,x,\,b = 4\,y\) and \(c = 3\,z\)
\( \Rightarrow {\left( { – 5\,x + 4\,y + 3\,z} \right)^2}\)
Hence, the factors of \(25\,{x^2} + 16\,{y^2} + 9\,{z^2} – 40\,xy + 24\,yz – 30\,zx\) are \(\left( {4\,y + 3\,z – 5\,x} \right)\left( {4\,y + 3\,z – 5\,x} \right).\)

Q.3. Expand \({\left( {x – 3\,y} \right)^3}\) using standard algebraic identities.
Ans:
Given: \({\left( {x – 3\,y} \right)^3}\)
By using the identity: \({\left( {a – b} \right)^3} = {a^3} – {b^3} – 3\,ab\left( {a – b} \right)\)
By putting the value of \(a = x\) and \(b = 3\,y,\) we get,
\({\left( {x = 3\,y} \right)^3} = {x^3} – {\left( {3\,y} \right)^3} – 3{\left( x \right)^2}\left( {3\,y} \right) + 3\left( x \right){\left( {3\,y} \right)^2}\)
\( \Rightarrow {x^3} – 27\,{y^3} – 9\,{x^2}y + 27\,x{y^2}\)

Q.4. Factorise \(8\,{x^3} + 27\,{y^3} + 125\,{z^3} – 90\,xyz\) using standard algebraic identities.
Ans:
Given: \(8\,{x^3} + 27\,{y^3} + 125\,{z^3} – 90\,xyz\)
\( \Rightarrow {2^3}\,{x^3} + {3^3}\,{y^3} + {5^3}\,{z^3} – 90\,xyz\)
\( \Rightarrow {\left( {2\,x} \right)^3} + {\left( {3\,y} \right)^3} + {\left( {5\,z} \right)^3} – 3 \times 2\,x \times 3\,y \times 5\,z\)
By using the identity: \(\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} – ab – bc – ca} \right) = {a^3} + {b^3} + {c^3} – 3\,abc\)
By putting the values of \(a = 2\,x,\,b = 3\,y\) and \(c = 5\,z\)
\( \Rightarrow \left( {2x + 3y + 5z} \right)\left[ {{{\left( {2x} \right)}^2} + \left( {3{y^2}} \right) + {{\left( {5z} \right)}^2} – \left( {2x} \right)\left( {3y} \right) – \left( {3y} \right)\left( {5z} \right) – \left( {5z} \right)\left( {2x} \right)} \right]\)
\( \Rightarrow \left( {2\,x + 3\,y + 5\,z} \right)\left( {4\,{x^2} + 9\,{y^2} + 25\,{z^2} – 6\,xy – 15\,yz – 10\,zx} \right)\)

Q.5. Factorise \(\left( {{x^4} – 1} \right)\) using standard algebraic identities.
Ans:
Given: \(\left( {{x^4} – 1} \right)\)
It can be written as \(\left[ {{{\left( {{x^2}} \right)}^2} – 1} \right]\)
By using the standard identity: \(\left( {a + b} \right)\left( {a – b} \right) = {a^2} – {b^2},\) we get,
\(\left( {{x^4} – 1} \right) = \left( {{x^2} – 1} \right)\left( {{x^2} + 1} \right)\)
\( = \left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)\) (Again, by using the identity: \(\left( {a + b} \right)\left( {a – b} \right) = {a^2} – {b^2}\))
Hence, the factors of \(\left( {{x^4} – 1} \right)\) are \(\left( {x – 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right).\)

Q.6. The area of a square is \(9\,{x^2} + 12\,x + 4.\) Find the measure of the side of the square.
Ans:
Given: Area of the square is \(9\,{x^2} + 12\,x + 4.\)
We know that area of the square is \({a^2},\) with side \(a.\)
So, \({a^2} = 9\,{x^2} + 12\,x + 4.\)
By using the identity: \({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)
\( \Rightarrow {a^2} = {3^2}\,{x^2} + 12\,x + {2^2}\)
\( \Rightarrow {a^2} = {\left( {3\,x} \right)^2} + 2\left( {3\,x} \right)\left( 2 \right) + {2^2}\)
\( \Rightarrow {a^2} = {\left( {3\,x + 2} \right)^2}\)
\( \Rightarrow a = 3\,x + 2\)
Hence, the side of the square is \(\left( {3\,x + 2} \right)\) units.

Q.7. Find the value of \(297 \times 303\) by using the standard algebraic identities.
Ans:
Given: \(297 \times 303\)
It can be written as \(\left( {300 – 3} \right) \times \left( {300 + 3} \right)\)
By using the identity: \(\left( {a + b} \right)\left( {a – b} \right) = {a^2} – {b^2}\)
Replace the value of \(a = 300\) and \(b = 3\)
\(\left( {300 – 3} \right) \times \left( {300 + 3} \right) = {\left( {300} \right)^2} – {3^2}\)
\( = 90000 – 9\)
\( = 89991\)
Hence, the value of \(297 \times 303 = 89991.\)

Summary

In this article, we have discussed what algebraic identities are and how much they are important in mathematics for solving any problems. We also discussed some standard identities and their proofs. Moreover, these identities form the basis for all the algebraic formulas. In the end, we have discussed algebraic identities with examples which will help you to understand this topic.

Frequently Asked Questions (FAQs) on Algebraic Identities

Q1. How do you verify the algebraic identities?
Ans: The algebraic identities are verified using the substitution method. In this method, substitute the values for the variables and perform the arithmetic operation. Another method to verify the algebraic identity is the activity method. In this method, you would need a prerequisite knowledge of Geometry, and some materials are needed to prove the identity.

Q2. What is the difference between algebraic identities and expressions?
Ans: Algebraic identity is the equality that cannot be changed for any values of the variable. An algebraic expression consists of variables, numbers, and mathematical operators. The value of an algebraic expression will change if the values of variables are changed.

Q3. How do you memorise algebraic identity?
Ans: Algebraic identities can be easily memorised by visualising them as squares or rectangles.
They can also be remembered by the factorised forms rather than the simplified forms.

Q4. What is standard identity?
Ans: The standard identity is the quality in which the left-hand side and right-hand side of the algebraic equation are true for all the values of the variables.

Q5. What all are the algebraic identities in Maths?
Ans: Four standard algebraic identities are listed below:
Identity-1: Algebraic Identity of Square of Sum of Two Terms
\({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)
Identity-2: Algebraic Identity of Square of Difference of Two Terms
\({\left( {a + b} \right)^2} = {a^2} + 2\,ab + {b^2}\)
Identity-3: Algebraic Identity of Difference of Two Squares
\(\left( {a + b} \right)\left( {a – b} \right) = {a^2} – {b^2}\)
Identity-4: Algebraic Identity \(\left( {x + a} \right)\left( {x + b} \right)\)
\(\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\)