SSC CGL 2027 Algebraic Identities and Simplification Practice

Algebraic Identities and Simplification are important topics in the Quantitative Aptitude section of SSC CGL 2027. These concepts help candidates solve mathematical expressions quickly, reduce lengthy calculations and improve accuracy in competitive examinations.

In SSC CGL, questions may involve algebraic formulas, expansion of expressions, factorisation, squares, cubes and numerical simplification. Learning the standard identities and practising their applications can help you save valuable time during the examination.

This article covers important algebraic identities, solved examples, shortcut methods and practice MCQs for SSC CGL Tier 1 and Tier 2 preparation.

1. Important Algebraic Identities for SSC CGL 2027

Memorise the following identities and understand when to apply each formula.

• Square of a sum: (a + b)² = a² + 2ab + b²
• Square of a difference: (a − b)² = a² − 2ab + b²
• Difference of squares: a² − b² = (a − b)(a + b)
• Product of sum and difference: (a + b)(a − b) = a² − b²
• Product of two binomials: (x + a)(x + b) = x² + (a + b)x + ab
• Cube of a sum: (a + b)³ = a³ + 3a²b + 3ab² + b³
• Cube of a difference: (a − b)³ = a³ − 3a²b + 3ab² − b³
• Sum of cubes: a³ + b³ = (a + b)(a² − ab + b²)
• Difference of cubes: a³ − b³ = (a − b)(a² + ab + b²)
• Square of three terms: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

2. Solved Examples on Algebraic Identities

Example 1: Find the value of 103² using an algebraic identity.

Solution:

103² = (100 + 3)²

Using (a + b)² = a² + 2ab + b²:

103² = 100² + 2 × 100 × 3 + 3²

= 10,000 + 600 + 9

Answer: 10,609

Example 2: Calculate 98² without lengthy multiplication.

Solution:

98² = (100 − 2)²

= 100² − 2 × 100 × 2 + 2²

= 10,000 − 400 + 4

Answer: 9,604

Example 3: Simplify (x + 5)² − (x − 5)².

Solution:

(x + 5)² = x² + 10x + 25

(x − 5)² = x² − 10x + 25

Subtracting the second expression from the first:

= x² + 10x + 25 − x² + 10x − 25

= 20x

Answer: 20x

3. Important Shortcuts for Squares

Algebraic identities are especially useful when calculating squares of numbers close to 100, 50 or another convenient base.

Example 4: Find 49².

Solution:

49² = (50 − 1)²

= 50² − 2 × 50 × 1 + 1²

= 2,500 − 100 + 1

Answer: 2,401

Example 5: Find 51 × 49.

Solution:

51 × 49 = (50 + 1)(50 − 1)

Using (a + b)(a − b) = a² − b²:

= 50² − 1²

= 2,500 − 1

Answer: 2,499

4. Factorisation Using Algebraic Identities

Factorisation means expressing an algebraic expression as a product of simpler factors. It can help candidates simplify expressions and solve algebra-based questions efficiently.

Example 6: Factorise x² − 81.

Solution:

x² − 81 = x² − 9²

Using a² − b² = (a − b)(a + b):

Answer: (x − 9)(x + 9)

Example 7: Factorise x² + 7x + 12.

Solution:

Find two numbers whose sum is 7 and product is 12.

Those numbers are 3 and 4.

x² + 7x + 12 = x² + 3x + 4x + 12

= x(x + 3) + 4(x + 3)

Answer: (x + 3)(x + 4)

5. Simplification Using Algebraic Formulas

Simplification questions may combine multiple operations, fractions, brackets and algebraic expressions. Follow the correct order of operations and apply identities wherever possible.

Example 8: Simplify (a + b)² − (a − b)².

Solution:

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab + b²

Subtracting:

= a² + 2ab + b² − a² + 2ab − b²

= 4ab

Answer: 4ab

Example 9: If x + 1/x = 5, find x² + 1/x².

Solution:

(x + 1/x)² = x² + 2 + 1/x²

25 = x² + 2 + 1/x²

x² + 1/x² = 25 − 2

Answer: 23

Example 10: If a − b = 4 and ab = 5, find a² + b².

Solution:

(a − b)² = a² − 2ab + b²

16 = a² + b² − 10

a² + b² = 16 + 10

Answer: 26

6. Cube Identities and Their Applications

Cube identities are useful in questions involving larger algebraic expressions and factorisation.

Example 11: Expand (x + 2)³.

Solution:

(x + 2)³ = x³ + 3x²(2) + 3x(2²) + 2³

= x³ + 6x² + 12x + 8

Answer: x³ + 6x² + 12x + 8

Example 12: Factorise x³ − 27.

Solution:

x³ − 27 = x³ − 3³

Using a³ − b³ = (a − b)(a² + ab + b²):

Answer: (x − 3)(x² + 3x + 9)

7. Common Mistakes to Avoid

• Forgetting the middle term 2ab when expanding (a + b)² or (a − b)².
• Writing a² − b² as (a − b)² instead of (a − b)(a + b).
• Making sign errors while subtracting expressions inside brackets.
• Confusing the formulas for the sum and difference of cubes.
• Ignoring the order of operations in simplification questions.
• Expanding a lengthy expression when a direct identity can solve it quickly.
• Using an identity without checking whether the expression matches its structure.

8. Practice Questions for SSC CGL 2027

Try these multiple-choice questions to test your understanding of algebraic identities and simplification.

Question 1: Find the value of 102².

• A) 10,204
• B) 10,404
• C) 10,304
• D) 10,504

Question 2: Simplify (a + b)² − (a − b)².

• A) 2ab
• B) 4ab
• C) 2a² + 2b²
• D) a² − b²

Question 3: Factorise x² − 64.

• A) (x − 8)²
• B) (x + 8)²
• C) (x − 8)(x + 8)
• D) (x − 4)(x + 16)

Question 4: If x + 1/x = 4, find x² + 1/x².

• A) 12
• B) 14
• C) 16
• D) 18

Question 5: Expand (x + 3)².

• A) x² + 3x + 9
• B) x² + 6x + 9
• C) x² + 9x + 6
• D) x² − 6x + 9

9. Answers with Explanations

1. B) 10,404
102² = (100 + 2)² = 10,000 + 400 + 4 = 10,404.

2. B) 4ab
Expanding both squares and subtracting gives 4ab.

3. C) (x − 8)(x + 8)
x² − 64 = x² − 8² = (x − 8)(x + 8).

4. B) 14
(x + 1/x)² = x² + 2 + 1/x². Therefore, 16 = x² + 2 + 1/x², so x² + 1/x² = 14.

5. B) x² + 6x + 9
(x + 3)² = x² + 2 × x × 3 + 3² = x² + 6x + 9.

10. How to Prepare Algebra for SSC CGL 2027

Start by memorising the standard square, cube and factorisation identities. Practise five to ten questions daily and focus on recognising the correct formula before beginning lengthy calculations.

Once you are comfortable with the basic formulas, solve mixed simplification questions under timed conditions. Review your mistakes and practise the identities that cause difficulty.

Conclusion

Algebraic identities and simplification can make many SSC CGL 2027 Quantitative Aptitude questions easier and faster to solve. A strong understanding of squares, cubes, factorisation and algebraic relationships helps improve both calculation speed and accuracy.

Revise the formulas regularly and practise a variety of questions to strengthen your preparation for SSC CGL Tier 1 and Tier 2.