Algebraic Identities: Definition, Questions, Formula, Examples

Algebraic Identities of Two Variables (a, b) – Formulas, Examples, and Applications

Algebraic identities involving two variables are the most commonly used formulas in mathematics. These identities form the foundation of algebra and are widely used in simplification, expansion, factorisation, and solving equations. Understanding these algebraic identities formulas is essential for both school exams and competitive exams.

Important Algebraic Identities of Two Variables

The key algebraic identities involving variables $a$ and $b$ are:

• $(a+b)^2 = a^2 + 2ab + b^2$
• $(a-b)^2 = a^2 - 2ab + b^2$
• $a^2 + b^2 = (a+b)^2 - 2ab = (a-b)^2 + 2ab$
• $a^2 - b^2 = (a+b)(a-b)$
• $(x+a)(x+b) = x^2 + (a+b)x + ab$
• $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a+b)$
• $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a-b)$
• $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
• $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

Key insights:

• These identities help in quick expansion and factorisation
• They are heavily used in polynomial and quadratic expressions
• Frequently asked in CBSE, CUET, JEE, and aptitude exams

Algebraic Identities of Three Variables (a, b, c) – Complete Formula List

Algebraic identities with three variables are slightly more advanced and are commonly used in higher-level algebra problems.

Important Identities of Three Variables

• $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
• $a^2 + b^2 + c^2 = (a+b+c)^2 - 2ab - 2bc - 2ca$
• $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$
• If $a+b+c = 0$, then $a^3 + b^3 + c^3 = 3abc$

Where these are used:

• Simplifying multi-variable algebraic expressions
• Solving higher-order algebra problems
• Frequently used in Olympiad and competitive exams

Algebraic Identities Used for Factorisation (Shortcut Methods)

Algebraic identities play a major role in factorisation, which is essential for solving equations quickly.

Important Factorisation Identities

• $a^2 - b^2 = (a+b)(a-b)$
• $x^2 + (a+b)x + ab = (x+a)(x+b)$
• $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
• $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

Why these identities matter:

• Help break complex expressions into simpler factors
• Reduce calculation time significantly
• Essential for solving quadratic and cubic equations
• Widely used in board exams and entrance tests

Proof of Standard Algebraic Identities (Conceptual Understanding)

Understanding the proof of algebraic identities helps build strong conceptual clarity. Instead of memorising formulas blindly, learning their derivation ensures better application in exams.

Proof of $(x + a)(x + b) = x^2 + x(a + b) + ab$

This identity can be understood using multiplication:

• Multiply each term:
  $x(x+b) + a(x+b)$
• Expand:
  $x^2 + xb + ax + ab$
• Combine like terms:
  $x^2 + x(a+b) + ab$

Conclusion:

• This identity shows how two linear expressions multiply into a quadratic expression
• Widely used in factorisation and quadratic equations

Proof of $(a + b)^2 = a^2 + 2ab + b^2$

Step-by-step expansion:

• $(a+b)(a+b)$
• Multiply each term:
  $a(a+b) + b(a+b)$
• Expand:
  $a^2 + ab + ab + b^2$
• Combine like terms:
  $a^2 + 2ab + b^2$

Key idea:

• The middle term $2ab$ comes from adding $ab + ab$
• This identity is one of the most frequently used in algebra

Proof of $(a + b)(a - b) = a^2 - b^2$

Step-by-step expansion:

• $(a+b)(a-b)$
• Multiply:
  $a(a-b) + b(a-b)$
• Expand:
  $a^2 - ab + ab - b^2$
• Simplify:
  $a^2 - b^2$

Key observation:

• The middle terms cancel out
• This is why it is called the difference of squares

Proof of $(a - b)^2 = a^2 - 2ab + b^2$

Step-by-step expansion:

• $(a-b)(a-b)$
• Multiply:
  $a(a-b) - b(a-b)$
• Expand:
  $a^2 - ab - ab + b^2$
• Simplify:
  $a^2 - 2ab + b^2$

Important point:

• The negative sign affects the middle term
• This is a common area where students make mistakes

Important Algebraic Identities Table (Formulas)

These formulas are essential for exams like CBSE Class 9 and 10, CUET, JEE Main, and IPMAT. Whether you are a beginner or preparing for competitive exams, mastering these identities is crucial. The table below presents all the important algebraic identities formulas in a simple, organised, and exam-focused format.

Application of Algebraic Identities

• Simplifying Expressions: We can reduce the complexity of algebraic expressions and factorise large polynomials with the help of algebraic identities.

• Solving Equations: We can solve quadratic and higher-degree equations by using the identities.

• Geometric Calculations: These identities also help us calculate the areas and volumes of geometric shapes easily.

• Probability and Statistics: We can derive formulas for variance and standard deviation and also simplify complex expressions using the identities.

• Trigonometry: Using these identities we can also simplify trigonometric expressions and prove trigonometric formulas.

• Computer Science: We can optimize algorithms and design error-detecting or correcting codes with the help of these identities.

• Economics and Finance: We can also derive and simplify formulas for compound interest and solve optimization problems using algebraic identities.

These applications demonstrate the versatility and importance of algebraic identities across various fields.

How to Solve Questions Using Algebraic Identities Quickly

This section focuses on how to solve algebraic identities questions quickly, using structured steps, avoiding common errors, and applying exam-oriented strategies.

Step-by-Step Method to Apply Algebraic Identities

A systematic approach helps you identify the correct identity and apply it without confusion.

Follow this method:

Practical example: Expand $(2x+7)^2$

• Identify: $(a+b)^2$
• Apply: $(2x)^2 + 2(2x)(7) + 7^2$
• Result: $4x^2 + 28x + 49$

Key takeaway:

• Always recognise the pattern first, then apply the formula

Common Mistakes to Avoid in Algebraic Identities Questions

Even if you know the formulas, small mistakes can lead to wrong answers. These errors are very common in exams.

Most frequent mistakes:

How to avoid these mistakes:

• Revise formulas regularly
• Practice step-by-step solving
• Double-check signs and coefficients
• Verify the final expression

Algebraic Identities Tricks, Shortcuts, and Memory Techniques

Learning algebraic identities is not just about memorisation. Smart tricks and shortcuts can help you remember formulas easily and apply them quickly in exams.

Easy Tricks to Remember Algebraic Identities Formulas

Memorising identities becomes easier when you understand patterns instead of rote learning.

Simple memory techniques:

• Use pattern logic
  • $(a+b)^2$ → first square + twice product + second square
  • $(a-b)^2$ → same pattern but middle term negative
• Visualise structure
  • First term squared
  • Middle term as $2ab$
  • Last term squared
• Learn in groups
  • Square identities together
  • Cube identities together
  • Factorisation identities together
• Repeat through practice
  • Write formulas daily
  • Apply them in questions

Quick recall trick:

• Think: square → 3 terms
• Cube → 4 terms
• Difference of squares → 2 terms

Shortcut Methods for Expansion and Factorisation

Shortcuts help reduce calculation time and are extremely useful in exams.

Expansion shortcuts:

• $(a+b)^2$ → directly write $a^2 + 2ab + b^2$
• $(a-b)^2$ → write $a^2 - 2ab + b^2$
• $(a+b)(a-b)$ → write $a^2 - b^2$

Factorisation shortcuts:

• Identify difference of squares
  • $x^2 - 16$ → $(x+4)(x-4)$
• Identify quadratic patterns
  • $x^2 + 5x + 6$ → $(x+2)(x+3)$
• Recognise cube identities
  • $a^3 + b^3$ → factor using standard identity

Key idea:

• Do not expand unnecessarily
• Always check if factorisation is faster

Exam-Oriented Tricks for Faster Problem Solving

In competitive exams, time is limited, so applying algebraic identities efficiently is crucial.

Exam-focused strategies:

• Convert numbers into identity form
  • $99^2 = (100-1)^2$
  • $101 \times 99 = (100+1)(100-1)$
• Look for hidden identities in questions
  • Many questions are designed to test recognition
• Avoid long multiplication
  • Always check if an identity can be applied
• Practice high-frequency questions
  • Focus on previous year exam patterns
• Use approximation where applicable
  • Helps in quick elimination in MCQs

Final takeaway:

• Speed comes from recognising patterns, not from calculation
• Algebraic identities tricks and shortcuts are essential for scoring high in exams like CBSE, CUET, JEE, and IPMAT

Mastering these strategies ensures that you can solve algebraic identities questions quickly, accurately, and with confidence under exam pressure.

Practice Questions/Solved Examples for Algebraic Identities

Q.1 If $x + y = \sqrt{3}$ and $x - y = \sqrt{2}$, the value of $8xy(x^2 + y^2)$ is:

A. $6$
B. $\sqrt{6}$
C. $5$
D. $\sqrt{5}$

Solution:

$x + y = \sqrt{3}$

Square both sides:
$(x+y)^2 = (\sqrt{3})^2$

$x^2 + y^2 + 2xy = 3 \quad ...(i)$

$x - y = \sqrt{2}$

Square both sides:
$(x-y)^2 = (\sqrt{2})^2$

$x^2 + y^2 - 2xy = 2 \quad ...(ii)$

Add (i) and (ii):
$(x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) = 3 + 2$

$2x^2 + 2y^2 = 5$

Divide both sides by 2:
$x^2 + y^2 = \frac{5}{2} \quad ...(iii)$

Substitute (iii) into (i):
$\frac{5}{2} + 2xy = 3$

Subtract $\frac{5}{2}$ from both sides:
$2xy = 3 - \frac{5}{2}$

Convert to common denominator:
$3 = \frac{6}{2}$

$2xy = \frac{6}{2} - \frac{5}{2} = \frac{1}{2}$

Divide both sides by 2:
$xy = \frac{1}{4}$

Now evaluate:
$8xy(x^2 + y^2)$

Substitute values:
$= 8 \times \frac{1}{4} \times \frac{5}{2}$

First simplify:
$8 \times \frac{1}{4} = 2$

Now:
$2 \times \frac{5}{2} = 5$

Correct Option: C

Q.2 If $(a+b-6)^2 + a^2 + b^2 + 1 + 2b = 2ab + 2a$, the value of $a$ is:

A. $7$
B. $6$
C. $3.5$
D. $2.5$

Solution:

Given:
$(a+b-6)^2 + a^2 + b^2 + 1 + 2b = 2ab + 2a$

Bring all terms to LHS:
$(a+b-6)^2 + a^2 + b^2 + 1 + 2b - 2ab - 2a = 0$

Now rewrite:
$a^2 + b^2 - 2ab = (a-b)^2$

Group terms:
$(a+b-6)^2 + (a-b-1)^2 = 0$

If sum of squares = 0 → each = 0

$a+b-6 = 0 \Rightarrow a+b=6 \quad ...(i)$

$a-b-1 = 0 \Rightarrow a-b=1 \quad ...(ii)$

Add (i) and (ii):
$a+b + a-b = 6 + 1$

$2a = 7$

$a = \frac{7}{2} = 3.5$

Correct Option: C

Q.3 If $a^2 + 13b^2 + c^2 - 4ab - 6bc = 0$, then $a:b:c$ is:

A. $1:2:3$
B. $2:1:3$
C. $2:3:1$
D. $3:2:1$

Solution:

$a^2 + 13b^2 + c^2 - 4ab - 6bc = 0$

Split $13b^2$:
$a^2 - 4ab + 4b^2 + 9b^2 + c^2 - 6bc = 0$

Group:
$(a^2 - 4ab + 4b^2) + (c^2 - 6bc + 9b^2) = 0$

Write as squares:
$(a-2b)^2 + (c-3b)^2 = 0$

So,
$a-2b=0 \Rightarrow a=2b$

$c-3b=0 \Rightarrow c=3b$

Thus: $a:b:c = 2:1:3$

Correct Option: B

Q.4 If $(a + \frac{1}{a})^2 = 3$, find $a^2 + \frac{1}{a^2}$

Options:
A. $0$
B. $1$
C. $2$
D. $3$

Solution:
Expand: $a^2 + 2 + \frac{1}{a^2} = 3$
$a^2 + \frac{1}{a^2} = 3 - 2 = 1$

Correct Option: B

Q.5 If $(a - 2) + \frac{1}{a + 2} = -1$, find $(a+2)^2 + \frac{1}{(a+2)^2}$

Options:
A. $7$
B. $11$
C. $23$
D. $27$

Solution:
Given equation:
$(a - 2) + \frac{1}{a + 2} = -1$
Add $4$ to both sides:

Left side:
$(a - 2) + 4 = a + 2$

So equation becomes:
$(a + 2) + \frac{1}{a + 2} = -1 + 4$

Right side:
$-1 + 4 = 3$

Thus, $(a + 2) + \frac{1}{a + 2} = 3$
Let $x = a + 2$

Then equation becomes:
$x + \frac{1}{x} = 3$
Square both sides:

$\left(x + \frac{1}{x}\right)^2 = 3^2$
Apply identity:
$(x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2$

So, $x^2 + \frac{1}{x^2} + 2 = 9$
Subtract 2 from both sides:

$x^2 + \frac{1}{x^2} = 9 - 2$

$x^2 + \frac{1}{x^2} = 7$
Substitute back $x = a + 2$:

$(a+2)^2 + \frac{1}{(a+2)^2} = 7$

Correct Option: A

Q.6 If $a + b = 5$ and $a - b = 3$, find $a^2 + b^2$

Options:
A. $17$
B. $18$
C. $19$
D. $20$

Solution:
Given:
$a + b = 5 \quad ...(i)$
$a - b = 3 \quad ...(ii)$
Add (i) and (ii):

$(a+b) + (a-b) = 5 + 3$

$a + b + a - b = 8$

$2a = 8$
Divide by 2:

$a = 4$

Substitute into (i):

$4 + b = 5$

$b = 5 - 4 = 1$

Find required value:

$a^2 + b^2 = 4^2 + 1^2$

$= 16 + 1 = 17$

Correct Option: A

Q.7 If $a + b + c = 0$, find $a^3 + b^3 + c^3$

Options:
A. $abc$
B. $2abc$
C. $3abc$
D. $0$

Solution:
Use identity:

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$
Substitute $a + b + c = 0$:

$a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - bc - ca)$
Right side becomes: $0$

So, $a^3 + b^3 + c^3 - 3abc = 0$
Add $3abc$ to both sides:

$a^3 + b^3 + c^3 = 3abc$

Correct Option: C

Q.8 If $\frac{1}{p} + \frac{1}{q} = \frac{1}{p+q}$, find $p^3 - q^3$

Options:
A. $p-q$
B. $pq$
C. $1$
D. $0$

Solution:
Given:
$\frac{1}{p} + \frac{1}{q} = \frac{1}{p+q}$
Take LHS common denominator:

$\frac{q + p}{pq} = \frac{1}{p+q}$
Cross multiply: $(p+q)^2 = pq$
Expand LHS: $p^2 + q^2 + 2pq = pq$
Bring all terms to one side:

$p^2 + q^2 + 2pq - pq = 0$

$p^2 + q^2 + pq = 0 \quad ...(i)$
Use identity: $p^3 - q^3 = (p-q)(p^2 + q^2 + pq)$
Substitute from (i):

$p^3 - q^3 = (p-q)(0)$
$p^3 - q^3 = 0$

Correct Option: D

Q.9 If the square of the sum of three consecutive natural numbers exceeds the sum of their squares by 292, find the largest number

Options:
A. $5$
B. $6$
C. $3$
D. $8$

Solution:
Let numbers be:

$a = x-1$, $b = x$, $c = x+1$
Given:

$(a+b+c)^2 - (a^2 + b^2 + c^2) = 292$
Use identity:

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
Substitute:

$a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - (a^2 + b^2 + c^2) = 292$
Cancel terms:

$2ab + 2bc + 2ca = 292$
Divide by 2:

$ab + bc + ca = 146$
Substitute values:

$x(x-1) + x(x+1) + (x+1)(x-1) = 146$
Expand each term:

$x^2 - x + x^2 + x + x^2 - 1 = 146$
Simplify:

$3x^2 - 1 = 146$
Add 1 both sides:

$3x^2 = 147$
Divide by 3:

$x^2 = 49$
$x = 7$
Largest number = $x + 1 = 8$

Correct Option: D

Q.10 If $x + \frac{1}{x} = \sqrt{3}$, find $x^3 + \frac{1}{x^3}$

Options:
A. $\sqrt{3}$
B. $\frac{1}{\sqrt{3}}$
C. $0$
D. $1$

Solution:
Given:
$x + \frac{1}{x} = \sqrt{3}$
Cube both sides:

$\left(x + \frac{1}{x}\right)^3 = (\sqrt{3})^3$
Apply identity:

$(a+b)^3 = a^3 + b^3 + 3ab(a+b)$

So, $x^3 + \frac{1}{x^3} + 3(x \cdot \frac{1}{x})(x + \frac{1}{x}) = 3\sqrt{3}$
Simplify:

$x \cdot \frac{1}{x} = 1$

So,$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = 3\sqrt{3}$
Substitute given value:

$x^3 + \frac{1}{x^3} + 3\sqrt{3} = 3\sqrt{3}$
Subtract $3\sqrt{3}$ from both sides:

$x^3 + \frac{1}{x^3} = 0$

Correct Option: C

Related Quantitative Aptitude Topics

To strengthen your understanding of algebraic identities, it is important to explore closely related quantitative aptitude topics that build on the same concepts. These topics help improve problem-solving speed and accuracy in exams like CUET, JEE, IPMAT, and other aptitude-based tests.