In this guide
Algebra in SSC CGL is less about solving equations and more about recognising patterns. A question may look like it needs pages of expansion — if x + 1/x = 5, find x³ + 1/x³ — but with the right identity it takes three lines. Candidates who learn a dozen identities thoroughly and practise spotting them can turn algebra into one of their fastest sections.
The core identities
| Identity | Expansion |
|---|---|
| (a + b)² | a² + 2ab + b² |
| (a − b)² | a² − 2ab + b² |
| a² − b² | (a + b)(a − b) |
| (a + b)³ | a³ + b³ + 3ab(a + b) |
| (a − b)³ | a³ − b³ − 3ab(a − b) |
| a³ + b³ | (a + b)(a² − ab + b²) |
| a³ − b³ | (a − b)(a² + ab + b²) |
| (a + b + c)² | a² + b² + c² + 2(ab + bc + ca) |
| a³ + b³ + c³ − 3abc | (a + b + c)(a² + b² + c² − ab − bc − ca) |
Useful derived results
- (a + b)² + (a − b)² = 2(a² + b²)
- (a + b)² − (a − b)² = 4ab
- a² + b² + c² − ab − bc − ca = ½[(a − b)² + (b − c)² + (c − a)²]
The last one shows that if a² + b² + c² = ab + bc + ca, then a = b = c.
The x + 1/x family
If x + 1/x = k:
| Expression | Value |
|---|---|
| x² + 1/x² | k² − 2 |
| x³ + 1/x³ | k³ − 3k |
| x⁴ + 1/x⁴ | (k² − 2)² − 2 |
| (x − 1/x)² | k² − 4 |
If x − 1/x = k:
| Expression | Value |
|---|---|
| x² + 1/x² | k² + 2 |
| x³ − 1/x³ | k³ + 3k |
Worked example: If x + 1/x = 5, find x³ + 1/x³.
k³ − 3k = 125 − 15 = 110.
Worked example: If x + 1/x = 3, find x⁴ + 1/x⁴.
x² + 1/x² = 9 − 2 = 7; x⁴ + 1/x⁴ = 49 − 2 = 47.
Special case: x + 1/x = 2
Then x = 1, so any expression like x¹⁰⁰ + 1/x¹⁰⁰ = 2.
Special case: x + 1/x = √3
Then x³ + 1/x³ = (√3)³ − 3√3 = 3√3 − 3√3 = 0, and x⁶ = −1.
Using a + b + c = 0
Worked example: If a + b + c = 0, find (a² / bc) + (b² / ca) + (c² / ab).
Combine: (a³ + b³ + c³)/abc = 3abc/abc = 3.
Worked example: Find the value of (x − y)³ + (y − z)³ + (z − x)³ ÷ [(x − y)(y − z)(z − x)] — that is, the sum of the three cubes divided by the product.
The three terms add up to 0, so the numerator = 3(x − y)(y − z)(z − x). The value = 3.
The substitution method
When an identity is hard to spot, put simple values that satisfy the condition and test the options.
Worked example: If a + b = 10 and ab = 21, find a³ + b³.
a³ + b³ = (a + b)³ − 3ab(a + b) = 1,000 − 630 = 370.
Check with values: a = 3, b = 7 satisfy both; 27 + 343 = 370. ✓
Factorisation quick checks
- a² − b² = (a + b)(a − b): 97² − 3² = 100 × 94 = 9,400.
- 1,001 × 999 = (1,000 + 1)(1,000 − 1) = 1,000,000 − 1 = 9,99,999.
Common traps
| Trap | Correct approach |
|---|---|
| Sign errors in (a − b)³ | Write the identity before substituting |
| Using x + 1/x formulas for x − 1/x | Check which one is given |
| Expanding when an identity fits | Look for a pattern first |
Practice
- If x + 1/x = 4, find x² + 1/x².
- If x − 1/x = 2, find x³ − 1/x³.
- If a + b = 7 and ab = 12, find a² + b².
- If a + b + c = 0, find a³ + b³ + c³ when abc = 5.
- Find 105² − 95².
- If x + 1/x = 2, find x⁵⁰ + 1/x⁵⁰.
Answers: 1. 14. 2. 14. 3. 25. 4. 15. 5. 2,000. 6. 2.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
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