In this guide
Algebra questions in SSC CHSL are rarely about long derivations. Most can be answered in under a minute, either by recognising an identity or by substituting a simple value and checking the options. Candidates who find algebra slow are usually expanding everything from scratch. Candidates who find it fast have seen the same dozen patterns so many times that they spot them at once.
This guide covers those patterns: the core identities and where they come from, the x + 1/x family, the a + b + c = 0 result, linear equations, and value putting as a safety net.
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) |
Why they hold: each is just multiplication. (a + b)² = (a + b)(a + b) gives a², ab, ba and b², and the two middle terms combine to 2ab. You never need to memorise a line you can rebuild in ten seconds, but you do need to recognise the patterns on sight.
Rearrangements you will use constantly:
- a² + b² = (a + b)² − 2ab = (a − b)² + 2ab.
- a³ + b³ = (a + b)³ − 3ab(a + b).
- a² + b² + c² = (a + b + c)² − 2(ab + bc + ca).
The x + 1/x family
If x + 1/x = k, then:
- x² + 1/x² = k² − 2.
- x³ + 1/x³ = k³ − 3k.
- x⁴ + 1/x⁴ = (x² + 1/x²)² − 2.
If x − 1/x = k, then x² + 1/x² = k² + 2 and x³ − 1/x³ = k³ + 3k.
Why: square x + 1/x and the middle term is 2 × x × 1/x = 2, a plain number, so it can be subtracted off. Cube it and the extra part is 3 × x × 1/x × (x + 1/x) = 3k. These are the (a + b)² and (a + b)³ identities with ab = 1.
Special values. If x + 1/x = 2, then x = 1, so every power of x plus its reciprocal is also 2. If x + 1/x = −2, then x = −1.
When a + b + c = 0
Then a³ + b³ + c³ = 3abc.
Why: there is a general identity a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). If the first bracket is 0, the whole right side is 0.
Watch for it in disguise. In 7³ − 4³ − 3³ the numbers 7, −4 and −3 add to zero. In (x − y)³ + (y − z)³ + (z − x)³, the three brackets add to zero, so the expression equals 3(x − y)(y − z)(z − x).
Linear equations
For one variable, collect the x terms on one side and the numbers on the other. For two variables, eliminate one by adding or subtracting the equations, or substitute one into the other.
In word problems, let the unknown be x, write one equation that says exactly what the sentence says, and check your answer against the story before marking it.
Value putting
When the question gives a condition and asks for a value, choose numbers that satisfy the condition and evaluate. If a + b + c = 0, try a = 1, b = 1, c = −2. If x + 1/x = 2, take x = 1.
Why it's safe: if the answer is a fixed number for every value that satisfies the condition, it is that number for your chosen values too. Avoid values that make a denominator zero, and if two options still match, try a second set of values.
Eight worked questions
Q1. If a + b = 9 and ab = 20, find a² + b² and a³ + b³.
a² + b² = 81 − 40 = 41. a³ + b³ = 9³ − 3 × 20 × 9 = 729 − 540 = 189. Check with a = 4, b = 5: 16 + 25 = 41 and 64 + 125 = 189.
Q2. Find 103² − 97².
(103 + 97)(103 − 97) = 200 × 6 = 1,200.
Q3. If x + 1/x = 3, find x² + 1/x², x³ + 1/x³ and x⁴ + 1/x⁴.
x² + 1/x² = 9 − 2 = 7. x³ + 1/x³ = 27 − 9 = 18. x⁴ + 1/x⁴ = 7² − 2 = 47.
Q4. If x − 1/x = 4, find x² + 1/x² and x³ − 1/x³.
x² + 1/x² = 16 + 2 = 18. x³ − 1/x³ = 64 + 12 = 76.
Q5. Find the value of 7³ + (−4)³ + (−3)³.
7 − 4 − 3 = 0, so the value is 3 × 7 × (−4) × (−3) = 252. Direct check: 343 − 64 − 27 = 252.
Q6. If a + b + c = 10 and ab + bc + ca = 31, find a² + b² + c².
100 − 2 × 31 = 38. (One set that fits is 2, 3 and 5: 4 + 9 + 25 = 38.)
Q7. A pen costs ₹5 more than a pencil. Three pens and two pencils cost ₹65. Find the price of a pen.
Let a pencil cost p. Then 3(p + 5) + 2p = 65, so 5p = 50 and p = 10. A pen costs ₹15.
Q8. If a + b + c = 0, find a²/bc + b²/ca + c²/ab.
Put a = 1, b = 1, c = −2: 1/(−2) + 1/(−2) + 4/1 = 3.
Algebra confirms it: the expression is (a³ + b³ + c³)/abc = 3abc/abc = 3.
Common mistakes
| Mistake | Fix |
|---|---|
| (a + b)² written as a² + b² | The 2ab term is always there |
| x² + 1/x² = k² + 2 when x + 1/x = k | Plus gives minus 2; minus gives plus 2 |
| Missing the a + b + c = 0 pattern | Add the bases first whenever you see three cubes |
| Value putting that breaks the condition | Check your values satisfy the given equation |
| Sign slips when moving terms across | Write each step; change the sign when a term crosses the equals sign |
Practice
- If a − b = 5 and ab = 14, find a² + b².
- Find 56² − 44².
- If x + 1/x = 5, find x² + 1/x².
- Solve 5x + 3 = 3x + 15.
- The sum of three consecutive numbers is 72. Find the largest.
- Find 5³ + (−2)³ + (−3)³.
- If x + 1/x = 4, find x³ + 1/x³.
- Solve 2x + 3y = 12 and 3x − y = 7.
Answers:
- 53. (a − b)² + 2ab = 25 + 28.
- 1,200. (56 + 44)(56 − 44) = 100 × 12.
- 23. 25 − 2.
- x = 6. 2x = 12.
- 25. The middle number is 72 ÷ 3 = 24, so the numbers are 23, 24 and 25.
- 90. 5 − 2 − 3 = 0, so the value is 3 × 5 × (−2) × (−3) = 90. Check: 125 − 8 − 27 = 90.
- 52. 4³ − 3 × 4 = 64 − 12.
- x = 3, y = 2. From the second equation, y = 3x − 7. Then 2x + 9x − 21 = 12, so 11x = 33.
What to do next
- Write the identity table from memory, then derive the x + 1/x results from it.
- Do 20 identity questions in 15 minutes, noting which ones value putting would have solved faster.
- Practise spotting a + b + c = 0 in ten expressions with cubes.
- Use the same identities in simplification, and move on to geometry, where algebra shows up inside area and angle questions.
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 .
Get the next SSC CHSL guide by email
New guides every week. No spam, unsubscribe any time.