Theory & Concepts

SSC CGL Algebraic Identities Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Basic Algebraic Identities specifically designed for the SSC CGL 2025-26 pattern.

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Algebraic Identities form the core of Advanced Mathematics in SSC CGL. Questions that seem like long, complex arithmetic are actually basic algebra formulas in disguise. Memorizing these standard identity formulas and the x+1xx + \frac{1}{x} patterns will give you a massive speed advantage.

Learning path

  • Squares & Cubes Formulas
  • The Inverse Pattern
  • The 3-Variable Identity
  • 10 Exam-Level Examples

1. Basic Squares & Cubes

These standard identities are used endlessly to expand or factorize expressions.

Square Identities

(a+b)2=a2+b2+2ab(a+b)^2 = a^2 + b^2 + 2ab

(a−b)2=a2+b2−2ab(a-b)^2 = a^2 + b^2 - 2ab

a2−b2=(a+b)(a−b)a^2-b^2 = (a+b)(a-b)

(a+b)2−(a−b)2=4ab(a+b)^2 - (a-b)^2 = 4ab

Cube Identities

(a+b)3=a3+b3+3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b)

(a−b)3=a3−b3−3ab(a−b)(a-b)^3 = a^3 - b^3 - 3ab(a-b)

a3+b3=(a+b)(a2−ab+b2)a^3+b^3 = (a+b)(a^2-ab+b^2)

a3−b3=(a−b)(a2+ab+b2)a^3-b^3 = (a-b)(a^2+ab+b^2)

2. The x+1xx + \frac{1}{x} Pattern

In Tier 1, SSC will always ask at least one question from this framework. You should answer these mentally.

Inverse Formulas

If x+1x=kx + \frac{1}{x} = k, then:

1. Square: x2+1x2=k2−2x^2 + \frac{1}{x^2} = k^2 - 2

2. Cube: x3+1x3=k3−3kx^3 + \frac{1}{x^3} = k^3 - 3k

Conversely, if x−1x=kx - \frac{1}{x} = k:

3. Square: x2+1x2=k2+2x^2 + \frac{1}{x^2} = k^2 + 2

4. Cube: x3−1x3=k3+3kx^3 - \frac{1}{x^3} = k^3 + 3k

To reverse a square (i.e. finding x+1xx + \frac{1}{x} from x2+1x2x^2 + \frac{1}{x^2}), add 2 and take the square root.

3. The 3-Variable Master Formula

This is the most important formula for SSC CGL Tier 2 algebra.

The Core Formula

a3+b3+c3−3abca^3+b^3+c^3-3abc
=(a+b+c)(a2+b2+c2−ab−bc−ca)= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

The Zero Condition

If a+b+c=0a + b + c = 0, the right side of the equation becomes 0.

a3+b3+c3=3abca^3 + b^3 + c^3 = 3abc

4. 10 Solved examples

Question 01Exam Pattern
If \( x + \frac{1}{x} = 4 \), find the value of \( x^2 + \frac{1}{x^2} \).
14
16
18
12
Correct answer: a) 14

Solution

Step 1: The given value is \( k = 4 \).
Step 2: Use the shortcut formula for squares: \( x^2 + \frac{1}{x^2} = k^2 - 2 \).
Step 3: Substitute \( k = 4 \): \( 4^2 - 2 \).
Final calculation: \( 16 - 2 = 14 \).
Question 02Exam Pattern
If \( x - \frac{1}{x} = 3 \), find the value of \( x^2 + \frac{1}{x^2} \).
7
9
11
13
Correct answer: c) 11

Solution

Step 1: Since we start with a minus, the formula has a plus.
Step 2: \( x^2 + \frac{1}{x^2} = k^2 + 2 \).
Step 3: Substitute \( k = 3 \): \( 3^2 + 2 \).
Final calculation: \( 9 + 2 = 11 \).
Question 03Exam Pattern
If \( x + \frac{1}{x} = 5 \), find the value of \( x^3 + \frac{1}{x^3} \).
125
110
115
140
Correct answer: b) 110

Solution

Step 1: The given value is \( k = 5 \).
Step 2: Use the shortcut formula for cubes: \( x^3 + \frac{1}{x^3} = k^3 - 3k \).
Step 3: Substitute \( k = 5 \): \( 5^3 - 3(5) \).
Step 4: Calculate the values: \( 125 - 15 \).
Final calculation: Result = 110.
Question 04Exam Pattern
If \( a + b = 7 \) and \( ab = 10 \), find the value of \( a^3 + b^3 \).
133
343
210
153
Correct answer: a) 133

Solution

Step 1: Use the identity \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \).
Step 2: Substitute the given values: \( (7)^3 - 3(10)(7) \).
Step 3: Evaluate the cube: \( 7^3 = 343 \).
Step 4: Evaluate the product: \( 3 \times 10 \times 7 = 210 \).
Final calculation: \( 343 - 210 = 133 \).
Question 05Exam Pattern
If \( x+y+z = 9 \) and \( xy+yz+zx = 23 \), find \( x^2+y^2+z^2 \).
35
45
55
81
Correct answer: a) 35

Solution

Step 1: Use the identity \( (x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx) \).
Step 2: Substitute the knowns: \( (9)^2 = x^2+y^2+z^2 + 2(23) \).
Step 3: Simplify: \( 81 = x^2+y^2+z^2 + 46 \).
Final calculation: \( x^2+y^2+z^2 = 81 - 46 = 35 \).
Question 06Exam Pattern
Find the value of \( a^3+b^3+c^3 - 3abc \) when \( a=99, b=100, c=101 \).
900
0
300
600
Correct answer: a) 900

Solution

Step 1: When values are large and consecutive, use the modified identity:
Step 2: \( a^3+b^3+c^3-3abc = \frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] \).
Step 3: Calculate \( a+b+c = 99+100+101 = 300 \).
Step 4: Calculate the differences: \( (-1)^2 + (-1)^2 + (2)^2 = 1 + 1 + 4 = 6 \).
Final calculation: \( \frac{1}{2} \times 300 \times 6 = 150 \times 6 = 900 \).
Question 07Exam Pattern
If \( x-y = 2 \) and \( xy = 15 \), find the value of \( x^3 - y^3 \).
98
8
60
120
Correct answer: a) 98

Solution

Step 1: Use the identity \( x^3 - y^3 = (x-y)^3 + 3xy(x-y) \).
Step 2: Substitute the known values: \( (2)^3 + 3(15)(2) \).
Step 3: Evaluate the cube: \( 2^3 = 8 \).
Step 4: Evaluate the product: \( 3 \times 15 \times 2 = 90 \).
Final calculation: \( 8 + 90 = 98 \).
Question 08Exam Pattern
If \( a+b+c = 0 \), find the value of \( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \).
0
1
2
3
Correct answer: d) 3

Solution

Step 1: Take the LCM of the denominators, which is \( abc \).
Step 2: The numerator becomes \( a^3 + b^3 + c^3 \). The expression is \( \frac{a^3 + b^3 + c^3}{abc} \).
Step 3: Since \( a+b+c = 0 \), we know that \( a^3 + b^3 + c^3 = 3abc \).
Final calculation: \( \frac{3abc}{abc} = 3 \).
Question 09Exam Pattern
If \( x^4 + \frac{1}{x^4} = 47 \), find the value of \( x + \frac{1}{x} \) (assume \( x > 0 \)).
3
5
7
9
Correct answer: a) 3

Solution

Step 1: We must step down the powers by adding 2 and taking the square root.
Step 2: \( x^2 + \frac{1}{x^2} = \sqrt{47 + 2} = \sqrt{49} = 7 \).
Step 3: Step down again to find \( x + \frac{1}{x} \).
Step 4: \( x + \frac{1}{x} = \sqrt{7 + 2} = \sqrt{9} \).
Final calculation: The value is 3.
Question 010Exam Pattern
Find the value of \( (a+b)^2 - (a-b)^2 \) if \( ab = 25 \).
50
100
0
25
Correct answer: b) 100

Solution

Step 1: Expand both brackets or use the standard identity.
Step 2: The identity states that \( (a+b)^2 - (a-b)^2 = 4ab \).
Step 3: We are given that \( ab = 25 \).
Final calculation: \( 4 \times 25 = 100 \).

5. Strategy errors to avoid

!

The Squaring Slip

The most common mistake is thinking \\( (a+b)^2 = a^2 + b^2 \\). Never forget the middle term \\( 2ab \\)! This is why \\( x^2 + 1/x^2 \\) is \\( k^2 - 2 \\) and not just \\( k^2 \\).

!

Sign Confusion in Inverses

Remember: \\( x + 1/x \\) leads to a MINUS in the result (\\( k^2-2 \\)), but \\( x - 1/x \\) leads to a PLUS (\\( k^2+2 \\)). Students often swap these under exam pressure.

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Linear EquationsQuadratic EquationsPolynomials & Factorization