SSC CGL 2026 Maths: Important Formulas, Identities, Tricks and Questions

Algebra is a key scoring area for aspirants aiming to increase their marks in the SSC CGL 2026 exam. Many students struggle with algebraic identities, especially those involving squares and cubes, which are frequently asked in exams. 

Here we break down these concepts in a systematic way, helping you master formulas, understand coefficient-based variations, and solve problems quickly and accurately.

Why Algebraic Identities Matter in SSC CGL Exams

Algebraic identities are not just theoretical formulas; they are shortcut tools used to solve complex questions in seconds. In SSC CGL exams, questions based on expressions like x + (1/x) and x − (1/x)  often appear in tricky forms to test speed and conceptual clarity. Understanding how to transform and apply these identities helps reduce calculation time and improves accuracy in competitive exams.

1. Basic Algebraic Identities for Squares

Understanding the basic identities for squares is fundamental.

If x + (1/x) = k:

To find x² + (1/x²): k² − 2

Explanation: This identity comes from expanding (a + b)² = a² + b² + 2ab. When a = x and b = (1/x), the 2ab term simplifies to 2(x)(1/x) = 2. Therefore,
(x + (1/x))² = x² + (1/x²) + 2
Rearranging gives:
x² + (1/x²) = (x + (1/x))² − 2

To find x⁴ + (1/x⁴): Let x² + (1/x²) = t. Then,
x⁴ + (1/x⁴) = t² − 2

To find x⁸ + (1/x⁸): Let x⁴ + (1/x⁴) = u. Then,
x⁸ + (1/x⁸) = u² − 2

General Rule for Squares: When squaring a sum (a + b), the +2ab term is moved to the right-hand side, resulting in subtraction.

If x − (1/x) = k:

To find x² + (1/x²): k² + 2

Explanation: This identity comes from (a − b)² = a² + b² − 2ab. When a = x and b = (1/x), the −2ab term simplifies to −2(x)(1/x) = −2. Therefore,
(x − (1/x))² = x² + (1/x²) − 2
Rearranging gives:
x² + (1/x²) = (x − (1/x))² + 2

General Rule for Squares: When squaring a difference (a − b), the −2ab term is moved to the right-hand side, resulting in addition.

2. Relationship Between x + (1/x) and x − (1/x)

It is often necessary to convert between these two forms while solving algebraic identity questions.

• From x + (1/x) = k to x − (1/x):

• x − (1/x) = √(k² − 4)

• From x − (1/x) = k to x + (1/x):

• x + (1/x) = √(k² + 4)

• Memory Tip: To convert from x + (1/x) to x − (1/x), use −4 inside the square root. To convert from x − (1/x) to x + (1/x), use +4 inside the square root.

3. Handling Coefficients in Square Identities

When x and 1/x have coefficients, the identities need to be adjusted accordingly.

If ax + b/(cx) = k and you need (ax)² + [b/(cx)]²:

The result is:
k² − 2ab/c

Explanation: Using the identity
(A + B)² = A² + B² + 2AB
where A = ax and B = b/(cx),
2AB = 2(ax)·[b/(cx)] = 2ab/c
Therefore,
(ax)² + [b/(cx)]² = k² − 2ab/c

If ax − b/(cx) = k and you need (ax)² + [b/(cx)]²:

The result is:
k² + 2ab/c

Explanation: Using the identity
(A − B)² = A² + B² − 2AB
where A = ax and B = b/(cx),
2AB = 2(ax)·[b/(cx)] = 2ab/c
Therefore,
(ax)² + [b/(cx)]² = k² + 2ab/c

4. Basic Algebraic Identities for Cubes

Similar to square identities, cube identities are frequently used in SSC CGL algebra questions.

If x + (1/x) = k:

To find x³ + (1/x³):
x³ + (1/x³) = k³ − 3k

Explanation: This identity is derived from
(a + b)³ = a³ + b³ + 3ab(a + b)
Let a = x and b = (1/x). Then,
(x + (1/x))³ = x³ + (1/x³) + 3(x)(1/x)(x + (1/x))
Since (x)(1/x) = 1, this becomes:
(x + (1/x))³ = x³ + (1/x³) + 3(x + (1/x))
Given x + (1/x) = k,
k³ = x³ + (1/x³) + 3k
Therefore,
x³ + (1/x³) = k³ − 3k

If x − (1/x) = k:

To find x³ − (1/x³):
x³ − (1/x³) = k³ + 3k

Explanation: This identity is derived from
(a − b)³ = a³ − b³ − 3ab(a − b)
Let a = x and b = (1/x). Then,
(x − (1/x))³ = x³ − (1/x³) − 3(x)(1/x)(x − (1/x))
Since (x)(1/x) = 1, this becomes:
(x − (1/x))³ = x³ − (1/x³) − 3(x − (1/x))
Given x − (1/x) = k,
k³ = x³ − (1/x³) − 3k
Therefore,
x³ − (1/x³) = k³ + 3k

5. Handling Coefficients in Cube Identities

Coefficients also affect cube identities and must be incorporated into the middle term of the expansion.

If ax + b/(cx) = k and you need (ax)³ + [b/(cx)]³:

The result is:
k³ − 3k(ab/c)

Explanation: Using the identity
A³ + B³ = (A + B)³ − 3AB(A + B)
where A = ax and B = b/(cx),
AB = (ax) · [b/(cx)] = ab/c
Since A + B = k,
(ax)³ + [b/(cx)]³ = k³ − 3k(ab/c)

If ax − b/(cx) = k and you need (ax)³ − [b/(cx)]³:

The result is:
k³ + 3k(ab/c)

Explanation: Using the identity
A³ − B³ = (A − B)³ + 3AB(A − B)
where A = ax and B = b/(cx),
AB = (ax) · [b/(cx)] = ab/c
Since A − B = k,
(ax)³ − [b/(cx)]³ = k³ + 3k(ab/c)

6. Calculating Higher Powers (e.g., Power 6, Power 8)

To find higher powers, use a sequential approach.

• To find Power 6 (x⁶ + 1/x⁶):

• Method 1 (Square then Cube): Calculate x² + 1/x², then cube that result. (x²)³ = x⁶.

• Method 2 (Cube then Square): Calculate x³ + 1/x³, then square that result. (x³) ² = x⁶.

• To find Power 8 (x⁸ + 1/x⁸):

• Calculate x² + 1/x² (let this be t).

• Calculate x⁴ + 1/x⁴ (t² - 2, let this be u).

• Calculate x⁸ + 1/x⁸ (u² - 2). This involves a repeated squaring process.

7. Problem-Solving Applications

Let's apply these concepts to common problem types.

• Basic Applications:

• Given: x + 1/x = 3

• x² + 1/x² = 3² - 2 = 7

• x⁴ + 1/x⁴ = 7² - 2 = 47

• x³ + 1/x³ = 3³ - 3(3) = 27 - 9 = 18

• Given: x - 1/x = 4

• x² + 1/x² = 4² + 2 = 18

• x⁴ + 1/x⁴ = 18² - 2 = 322

• x³ - 1/x³ = 4³ + 3(4) = 64 + 12 = 76

• Handling Square Roots:

• Given: √x - 1/√x = √6

• Square both sides: (√x - 1/√x)² = (√6)²

• x + 1/x - 2(√x)(1/√x) = 6

• x + 1/x - 2 = 6

• x + 1/x = 8

• Now, to find x² + 1/x²: 8² - 2 = 64 - 2 = 62.

• This shows that two squaring operations are needed to reach x² + 1/x² from √x - 1/√x.

• From Quadratic Equation to x + 1/x:

• Given: x² - 5x + 1 = 0

• Method: Divide the entire equation by x (assuming x ≠ 0).

• x²/x - 5x/x + 1/x = 0/x

• x - 5 + 1/x = 0

• x + 1/x = 5.

• Once x + 1/x is known, you can find x³ + 1/x³ and other expressions.

• Coefficient Arrangement and Manipulation (Very Important):

• Problem: Given 5x + 1/3x = 5, find 9x² + 1/25x².

• Target Form: To obtain (3x)² + (1/5x)², we need to manipulate the given equation to 3x + 1/5x.

• Method: Multiply the given equation by 3/5.

• (3/5)(5x + 1/3x) = (3/5)(5)

• 3x + 1/5x = 3

• Now, apply the squaring rule with coefficients:

• (3x)² + (1/5x)² = 3² - 2(3)(1/5) = 9 - 6/5 = 9 - 1.2 = 7.8.

• Coefficient Arrangement for Cubes (Very, Very Important):

• Problem: Assume we need to find 64x³ + 1/64x³.

• Given: x + 1/16x = 3.

• Target Form: We need 4x + 1/4x.

• Method: Multiply the given equation by 4.

• 4(x + 1/16x) = 4(3)

• 4x + 1/4x = 12.

• Now, cube this expression using the coefficient rule:

• (4x)³ + (1/4x)³ = 12³ - 3(12)(4)(1/4) = 1728 - 36 = 1692.

• Crucial Point: If the actual question required a different base (e.g., 16x³ + 1/256x³) and you found 64x³ + 1/64x³, you might need to divide your result by the initial multiplication factor. (For example, dividing 1692 by 4 if needed for a different base.)

• Using Unit Digit for Large Numbers:

• When dealing with calculations like 102² - 2, you can use unit digits to quickly eliminate options.

• 102² ends in 2² = 4. So 102² - 2 will end in 4 - 2 = 2. This strategy is helpful in multiple-choice questions.

Common Mistakes to Avoid in Algebraic Identities

While solving SSC CGL Maths Algebra questions, students often make small but costly mistakes that affect accuracy and speed.

• Confusing identities like k² − 2 and k² + 2 due to incorrect identification of x + (1/x) and x − (1/x) forms.

• Incorrect handling of coefficients in expressions such as (ax)² + [b/(cx)]², especially while applying algebraic identities and forgetting to account for the cross term 2(ax) · [b/(cx)].

• Directly expanding expressions instead of first converting them into standard algebraic identity form.

• Skipping simplification steps, which leads to calculation errors in final answers.

• Not verifying intermediate steps in multi-power problems such as x⁴, x⁶, or higher powers like x⁸ + (1/x⁸).