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Algebraic Identities List for SSC: Squares, Cubes and a³ + b³ + c³ − 3abc

(a ± b)² and a² − b² • (a ± b)³ • a³ ± b³ • (a + b + c)² • a³ + b³ + c³ − 3abc • When a + b + c = 0

SSC Railway 📅 Last updated: 2026-10-23

An identity is an equation that holds for every value of its variables. SSC papers use a small set of them again and again, in two ways: to square or cube numbers near a round base quickly, and to find an expression such as a² + b² or a³ + b³ when only a + b and ab are given. In the second kind you never need to find a and b themselves.

Learn each identity in both directions. Read left to right, it expands a bracket; read right to left, it factorises an expression. Exam questions use the factorised direction more often than students expect, especially in fractions where a long numerator and denominator share a factor and cancel to something simple.

A good habit is to check every answer by substituting small numbers that fit the conditions. It takes a few seconds and catches sign errors, which are the most frequent slip with cube identities.

IdentityCommon use
(a + b)² = a² + 2ab + b²squares just above a base
(a − b)² = a² − 2ab + b²squares just below a base
a² − b² = (a + b)(a − b)simplifying differences
a² + b² = (a + b)² − 2abgiven a + b and ab
(a + b)² − (a − b)² = 4ablinking sum and difference

Worked example 1: 103²

(100 + 3)² = 10,000 + 600 + 9 = 10,609.

Worked example 2: 98²

(100 − 2)² = 10,000 − 400 + 4 = 9,604.

Worked example 3: a² + b² from a sum and a product

If a + b = 10 and ab = 21, then a² + b² = 100 − 42 = 58.

Identity
(a + b)³ = a³ + b³ + 3ab(a + b)
(a − b)³ = a³ − b³ − 3ab(a − b)
a³ + b³ = (a + b)(a² − ab + b²) = (a + b)³ − 3ab(a + b)
a³ − b³ = (a − b)(a² + ab + b²) = (a − b)³ + 3ab(a − b)

Worked example 4: a³ + b³

If a + b = 7 and ab = 10, then a³ + b³ = 343 − 3 × 10 × 7 = 133. Check: a = 2 and b = 5 give 8 + 125 = 133.

Worked example 5: a³ − b³

If a − b = 3 and ab = 10, then a³ − b³ = 27 + 90 = 117. Check: a = 5 and b = 2 give 125 − 8 = 117.

The factorised forms are useful for fractions. When a numerator is a³ + b³ and the denominator is a² − ab + b², the fraction simplifies to a + b.

Worked example 6: a fraction that collapses

(0.75³ + 0.25³) ÷ (0.75² − 0.75 × 0.25 + 0.25²) = 0.75 + 0.25 = 1.

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

Worked example 7: sum of squares

If a + b + c = 9 and ab + bc + ca = 26, then a² + b² + c² = 81 − 52 = 29.

The second identity has a famous consequence. If a + b + c = 0, the right side is zero, so a³ + b³ + c³ = 3abc. Questions often hide this by giving three numbers or expressions that add to zero.

Worked example 8: 2, 3 and −5

2 + 3 + (−5) = 0, so 2³ + 3³ + (−5)³ = 8 + 27 − 125 = −90.

And 3abc = 3 × 2 × 3 × (−5) = −90, exactly as the identity predicts.

  • ✘ Writing (a + b)² as a² + b², dropping the 2ab term.
  • ✘ Using +3ab(a − b) for a³ − b³ with the wrong sign. Check with small numbers.
  • ✘ Applying a³ + b³ + c³ = 3abc when a + b + c is not zero.
  • ✘ Solving for a and b when the identity gives the answer directly.

For questions given as x + 1/x, see x + 1/x problems. Identities also speed up simplification, and the formula list is in the Quant cheat code.

Q1. If a + b = 12 and ab = 35, what is a² + b²? easy

✅ Answer: (d) 74

💡 Explanation: a² + b² = (a + b)² − 2ab = 144 − 70 = 74. Check with a = 5 and b = 7: 25 + 49 = 74.

Q2. If a + b = 5 and ab = 6, what is a³ + b³? easy

✅ Answer: (b) 35

💡 Explanation: a³ + b³ = (a + b)³ − 3ab(a + b) = 125 − 90 = 35. Check with a = 2 and b = 3: 8 + 27 = 35.

Q3. If a − b = 4 and ab = 12, what is a³ − b³? medium

✅ Answer: (a) 208

💡 Explanation: a³ − b³ = (a − b)³ + 3ab(a − b) = 64 + 144 = 208. Check with a = 6 and b = 2: 216 − 8 = 208.

Q4. If a + b + c = 0 and abc ≠ 0, what is (a³ + b³ + c³) ÷ abc? medium

✅ Answer: (c) 3

💡 Explanation: When a + b + c = 0, the identity a³ + b³ + c³ − 3abc = (a + b + c)(…) gives a³ + b³ + c³ = 3abc, so the ratio is 3.

Q5. What is 999²? easy

✅ Answer: (b) 998,001

💡 Explanation: 999² = (1,000 − 1)² = 1,000,000 − 2,000 + 1 = 998,001.

Q6. If a + b + c = 12 and a² + b² + c² = 50, what is ab + bc + ca? hard

✅ Answer: (d) 47

💡 Explanation: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), so 144 = 50 + 2(ab + bc + ca) and ab + bc + ca = 47.

How can I check an identity answer quickly?

Pick small numbers that satisfy the given conditions and substitute them. If a + b = 5 and ab = 6, then a = 2 and b = 3 work, so any expression in a and b can be evaluated directly to confirm the identity's answer.

Is a + b + c = 0 the only way a³ + b³ + c³ can equal 3abc?

No. It also holds when a = b = c, because then a² + b² + c² − ab − bc − ca = 0. Questions usually give one of these two conditions and expect you to recognise it.

All problems and solutions on this page are original. No outside facts are relied on; every numeric answer is recomputed by software before publishing.

How this page was prepared. Drafted with AI assistance (Claude (Anthropic AI model)) from the sources listed above. 15 calculations on this page were recomputed by software and matched. Automated checks also covered structure, internal links and duplicate text. Software checks cannot confirm facts. Reviewed by praveenkumar on 2026-10-23 against the sources listed above.

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