Algebraic Identities: Squares, Cubes, x + 1/x and Simplification
Algebraic Identities: Squares, Cubes, x + 1/x and Simplification
Algebra in NTPC is mostly identity work: given a sum and a product, or x + 1/x, find a higher power without solving for x. Learn the eight identities below as a toolkit, and almost every question becomes one or two lines of arithmetic.
1. Identity Box
| Identity | Expanded form |
|---|---|
| (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) = (a + b)³ − 3ab(a + b) |
| a³ − b³ | (a − b)(a² + ab + b²) = (a − b)³ + 3ab(a − b) |
| a³ + b³ + c³ − 3abc | (a + b + c)(a² + b² + c² − ab − bc − ca) |
| (a + b + c)² | a² + b² + c² + 2(ab + bc + ca) |
If a + b + c = 0, then a³ + b³ + c³ = 3abc
(a + b)² − (a − b)² = 4ab (a + b)² + (a − b)² = 2(a² + b²)
2. The x + 1/x Ladder
| Given | Next step |
|---|---|
| x + 1/x = k | x² + 1/x² = k² − 2 |
| x + 1/x = k | x³ + 1/x³ = k³ − 3k |
| x − 1/x = m | x² + 1/x² = m² + 2; x³ − 1/x³ = m³ + 3m |
| x² + 1/x² = p | x⁴ + 1/x⁴ = p² − 2 |
| Any two rungs | x⁵ + 1/x⁵ = (x² + 1/x²)(x³ + 1/x³) − (x + 1/x) |
Quick table for x + 1/x = k:
| k | x² + 1/x² | x³ + 1/x³ | x⁴ + 1/x⁴ |
|---|---|---|---|
| 2 | 2 | 2 | 2 |
| 3 | 7 | 18 | 47 |
| 4 | 14 | 52 | 194 |
| 5 | 23 | 110 | 527 |
| √3 | 1 | 0 | −1 |
Special value: x + 1/x = 2 forces x = 1, so every power sum equals 2. x + 1/x = √3 gives x⁶ = −1 type results: x³ + 1/x³ = 0.
✗ x + 1/x = 4, so x² + 1/x² = 16 | ✓ Subtract the middle term: 16 − 2 = 14
✗ x + 1/x = 3, so x³ + 1/x³ = 27 | ✓ Subtract 3k: 27 − 9 = 18
हिंदी नोट: x + 1/x = k दिया हो तो x² + 1/x² = k² − 2 और x³ + 1/x³ = k³ − 3k। a + b + c = 0 हो तो a³ + b³ + c³ = 3abc। ये तीन सूत्र आधे बीजगणित प्रश्न हल कर देते हैं।
Exam Pointer: Verified NTPC pattern: x + 1/x given, higher powers or their ratio asked (April 2016 CBT-1). Sum-and-product, a + b + c = 0 and surd-substitution questions had no verified NTPC shift in our check and are tagged syllabus-based; they appear regularly in other RRB and SSC papers built on the same syllabus.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: x + 1/x given, find higher powers
[PYQ: NTPC CBT-1 12-Apr-2016 Shift-2]
EXAM LEVEL
Q. If x + 1/x = 4, find x² + 1/x² and x³ + 1/x³.
Squaring, x² + 1/x² = 16 − 2 = 14. Cubing, x³ + 1/x³ = 64 − 3 × 4 = 52.
Answer: 14 and 52
EXAMATLAS LEVEL
Q. If x + 1/x = 3, find x⁴ + 1/x⁴ and x⁵ + 1/x⁵.
x² + 1/x² = 9 − 2 = 7, so x⁴ + 1/x⁴ = 49 − 2 = 47. Also x³ + 1/x³ = 27 − 9 = 18. Multiplying the second and third rungs, (x² + 1/x²)(x³ + 1/x³) = x⁵ + 1/x⁵ + x + 1/x, so x⁵ + 1/x⁵ = 7 × 18 − 3 = 123.
Answer: 47 and 123
Pattern 2: Sum and product given, find squares and cubes
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. If a + b = 7 and ab = 10, find a² + b² and a³ + b³.
a² + b² = 49 − 20 = 29. a³ + b³ = 343 − 3 × 10 × 7 = 343 − 210 = 133. (The numbers are 5 and 2: 25 + 4 = 29, 125 + 8 = 133.)
Answer: 29 and 133
EXAMATLAS LEVEL
Q. If a − b = 3 and a² + b² = 29, find ab, a + b (a, b positive) and a³ − b³.
2ab = (a² + b²) − (a − b)² = 29 − 9 = 20, so ab = 10. (a + b)² = 29 + 20 = 49, so a + b = 7. a³ − b³ = (a − b)³ + 3ab(a − b) = 27 + 90 = 117.
Answer: ab = 10, a + b = 7, a³ − b³ = 117
Pattern 3: Three variables: a + b + c = 0 and a³ + b³ + c³ − 3abc
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. If a + b + c = 0 and abc = 4, find a³ + b³ + c³.
When the sum is zero, a³ + b³ + c³ = 3abc = 12.
Answer: 12
EXAMATLAS LEVEL
Q. If a + b + c = 12 and a² + b² + c² = 50, find ab + bc + ca and a³ + b³ + c³ − 3abc.
2(ab + bc + ca) = 144 − 50 = 94, so ab + bc + ca = 47. Then a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) = 12 × (50 − 47) = 36.
Answer: 47 and 36
Pattern 4: Surd values: x = p + √q, find x + 1/x or x − 1/x
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. If x = 2 + √3, find x + 1/x and x² + 1/x².
1/x = 1/(2 + √3) = 2 − √3 (since (2 + √3)(2 − √3) = 1). So x + 1/x = 4 and x² + 1/x² = 16 − 2 = 14.
Answer: 4 and 14
EXAMATLAS LEVEL
Q. If x = √5 + 2, find x³ − 1/x³.
(√5 + 2)(√5 − 2) = 1, so 1/x = √5 − 2 and x − 1/x = 4. Then x³ − 1/x³ = 4³ + 3 × 4 = 64 + 12 = 76.
Answer: 76
Pattern 5: Simplification with identities (decimal fractions)
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Simplify (5.6³ + 4.4³)/(5.6² − 5.6 × 4.4 + 4.4²).
This is (a³ + b³)/(a² − ab + b²) = a + b = 5.6 + 4.4 = 10.
Answer: 10
EXAMATLAS LEVEL
Q. Simplify (1.25³ − 0.75³)/(1.25² + 1.25 × 0.75 + 0.75²) + (1.25² − 0.75²)/(1.25 − 0.75).
The first fraction is (a³ − b³)/(a² + ab + b²) = a − b = 0.5. The second is (a² − b²)/(a − b) = a + b = 2. Total = 2.5.
Answer: 2.5
60-Second Revision
- x + 1/x = k gives k² − 2, k³ − 3k, then square again for the fourth power.
- From a + b and ab you can reach a² + b², a³ + b³ and a − b.
- a + b + c = 0 means a³ + b³ + c³ = 3abc.
- For x = p + √q with p² − q = 1, 1/x is simply p − √q.
- Spot (a³ ± b³)/(a² ∓ ab + b²) patterns: the answer is a ± b.