This MCQ module is based on: Cubes, Cube Roots & Chapter Exercises
TOPIC 3 OF 22
Cubes, Cube Roots & Chapter Exercises
🎓 Class 8
Mathematics
CBSE
Theory
Ch 1 — Squares, Square Roots, Cubes and Cube Roots
⏱ ~40 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Cubes, Cube Roots & Chapter Exercises
Targeting Class 8 level in Number Theory, with Basic difficulty.
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§1.2 — Cubic Numbers↗
A cube is a solid where all six faces are equal squares. How many unit cubes stack into a cube of side 2? Of side 3? The answer gives us cubic numbers.
Definition — Perfect Cube
A perfect cube is the result of multiplying a number by itself three times: \(n \times n \times n = n^3\). Read as "n cubed".Examples: \(1^3=1,\ 2^3=8,\ 3^3=27,\ 4^3=64,\ 5^3=125, \ldots\)
Table of Perfect Cubes (1³ to 20³)
| n | n³ | n | n³ | n | n³ | n | n³ |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 6 | 216 | 11 | 1331 | 16 | 4096 |
| 2 | 8 | 7 | 343 | 12 | 1728 | 17 | 4913 |
| 3 | 27 | 8 | 512 | 13 | 2197 | 18 | 5832 |
| 4 | 64 | 9 | 729 | 14 | 2744 | 19 | 6859 |
| 5 | 125 | 10 | 1000 | 15 | 3375 | 20 | 8000 |
Units Digit Pattern for Cubes
Unlike squares, every digit 0–9 is possible as the last digit of a cube! Cubes cycle through all ten possible last digits: 0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9. Also, a cube cannot end in exactly two zeros — trailing zeros in cubes always come in groups of 3.
Perfect Cubes and Consecutive Odd Numbers
Just as squares are sums of consecutive odd numbers, each cube is a sum of consecutive odd numbers forming a specific group:
1 = 1³1³ = 1
3 + 5 = 2³2³ = 8
7 + 9 + 11 = 3³3³ = 27
13 + 15 + 17 + 19 = 4³4³ = 64
21 + 23 + 25 + 27 + 29 = 5³5³ = 125
31 + 33 + 35 + 37 + 39 + 41 = 6³6³ = 216
🔬 Challenge: What is 91+93+95+97+99+101+103+105+107+109?
This is a set of 10 consecutive odd numbers. Which cube does it represent? Find the sum without adding term by term!
The group for n³ starts at \((n-1)^2 + 1\)-th odd number and has n terms. Here there are 10 terms, so it represents 10³ = 1000.
Verify: average = (91+109)/2 = 100; sum = 10 × 100 = 1000 ✓
The Hardy–Ramanujan Number — Taxicab Numbers↗
1729 — The Hardy–Ramanujan Number
1729
Way 1: 1³ + 12³ = 1 + 1728 = 1729
Way 2: 9³ + 10³ = 729 + 1000 = 1729
Next taxicab numbers: 4104 and 13832
Way 2: 9³ + 10³ = 729 + 1000 = 1729
Next taxicab numbers: 4104 and 13832
🔬 Try This: Find the two cube representations of 4104 and 13832
4104: 4104 = 2³ + 16³ = 9³ + 15³
Check: 8 + 4096 = 4104 ✓ and 729 + 3375 = 4104 ✓
13832: 13832 = 2³ + 24³ = 18³ + 20³
Check: 8 + 13824 = 13832 ✓ and 5832 + 8000 = 13832 ✓
Cube Roots↗ — Finding ∛n
Definition — Cube Root
If \(y = x^3\), then \(x\) is the cube root of y, written \(x = \sqrt[3]{y}\).Examples: \(\sqrt[3]{8} = 2,\ \sqrt[3]{27} = 3,\ \sqrt[3]{1000} = 10\).
In general: \(\sqrt[3]{n^3} = n\).
Finding Cube Roots via Prime Factorisation
A number is a perfect cube if its prime factors can be split into three identical groups. The product of one group is the cube root.
📝 Find ∛3375
3375 = 3 × 3 × 3 × 5 × 5 × 5 = (3×5) × (3×5) × (3×5) = 15³
Three identical groups, each = 3 × 5 = 15
∴ ∛3375 = 15
📝 Is 500 a perfect cube?
500 = 2 × 2 × 5 × 5 × 5 = 2² × 5³
The factor 2 appears only twice — cannot make three equal groups for all factors.
500 is NOT a perfect cube.
Cube Roots of Large Numbers — Unit Digit Trick
Since cube unit digits are unique (each digit 0–9 maps to a unique cube digit), we can determine the units digit of the cube root from the units digit of the cube:
| Last digit of n | Last digit of n³ | Last digit of n | Last digit of n³ |
|---|---|---|---|
| 1 | 1 | 6 | 6 |
| 2 | 8 | 7 | 3 |
| 3 | 7 | 8 | 2 |
| 4 | 4 | 9 | 9 |
| 5 | 5 | 0 | 0 |
📝 Guess ∛1331 without factorisation
1331 is a 4-digit number. 10³=1000 and 20³=8000, so the cube root is a 2-digit number between 10 and 20.
Last digit of 1331 is 1. From the table, a cube ending in 1 must have cube root ending in 1.
Two-digit numbers ending in 1 between 10–20: only 11.
∴ ∛1331 = 11. Verify: 11³ = 11×11×11 = 1331 ✓
Successive Differences
For perfect squares, taking differences level by level gives all-equal values at level 2. What happens for cubes?
🔧 Cube Root Analyser
Enter any number to check if it's a perfect cube and find/estimate its cube root.
§1.3 — A Pinch of History
From Clay Tablets to Sanskrit Texts
The first known lists of perfect squares and cubes were compiled by the Babylonians around 1700 BCE on clay tablets — used for land measurement and architectural calculations.
In ancient Sanskrit, varga (वर्ग) meant both the geometric square figure and the square power, while ghana (घन) meant both the solid cube and the cube power. These terms were used in India from at least the 3rd century BCE.
Aryabhata (499 CE) wrote: "A square figure of four equal sides and the number representing its area are called varga."
The word "root" in mathematics comes from Sanskrit mūla (मूल, meaning "root of a plant" or "origin"). Arabic adopted it as jidhr, Latin as radix — giving us "radical" (√). Brahmagupta (628 CE) used the term pada (foot/basis) for square root.
Figure it Out — Cubes (Pages 16–17)
Q1. Find the cube roots of 27000 and 10648.
∛27000: 27000 = 27 × 1000 = 3³ × 10³ = (3×10)³ = 30³. ∴ ∛27000 = 30
∛10648: 10648 = 8 × 1331 = 2³ × 11³ = (2×11)³ = 22³. ∴ ∛10648 = 22
(Alternatively: 10648 ends in 8 → cube root ends in 2; estimate between 20³=8000 and 30³=27000 → try 22: 22³=10648 ✓)
∛10648: 10648 = 8 × 1331 = 2³ × 11³ = (2×11)³ = 22³. ∴ ∛10648 = 22
(Alternatively: 10648 ends in 8 → cube root ends in 2; estimate between 20³=8000 and 30³=27000 → try 22: 22³=10648 ✓)
Q2. What number must you multiply 1323 by to make it a perfect cube?
1323 = 3 × 441 = 3 × 3 × 147 = 3 × 3 × 3 × 49 = 3³ × 7².
The factor 7 appears twice. We need a third 7. Multiply by 7:
1323 × 7 = 9261 = 3³ × 7³ = (3×7)³ = 21³.
Answer: Multiply by 7.
The factor 7 appears twice. We need a third 7. Multiply by 7:
1323 × 7 = 9261 = 3³ × 7³ = (3×7)³ = 21³.
Answer: Multiply by 7.
Q3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
(i) False. Odd × odd × odd = odd. E.g., 3³ = 27 (odd).
(ii) False. 2³ = 8 ends in 8; also 12³ = 1728 ends in 8. The digit 8 IS possible.
(iii) False. Smallest 2-digit number is 10; 10³ = 1000 (4 digits). The cube of any 2-digit number has at least 4 digits.
(iv) True. Largest 2-digit number is 99; 99³ = 970299 (6 digits). Actually this is 6 digits. Let's check: 99³ = 970,299 — 6 digits, not 7. Corrected: False — the largest 2-digit cube is 99³ = 970,299 which has 6 digits.
(v) False. Cubes have an ODD number of factors only if they are perfect squares as well (i.e., sixth powers like 64=2⁶). In general, cubes do not have an odd number of factors. E.g., 8=2³ has factors 1,2,4,8 — 4 factors (even).
(ii) False. 2³ = 8 ends in 8; also 12³ = 1728 ends in 8. The digit 8 IS possible.
(iii) False. Smallest 2-digit number is 10; 10³ = 1000 (4 digits). The cube of any 2-digit number has at least 4 digits.
(iv) True. Largest 2-digit number is 99; 99³ = 970299 (6 digits). Actually this is 6 digits. Let's check: 99³ = 970,299 — 6 digits, not 7. Corrected: False — the largest 2-digit cube is 99³ = 970,299 which has 6 digits.
(v) False. Cubes have an ODD number of factors only if they are perfect squares as well (i.e., sixth powers like 64=2⁶). In general, cubes do not have an odd number of factors. E.g., 8=2³ has factors 1,2,4,8 — 4 factors (even).
Q4. You are told 1331 is a perfect cube. Guess its cube root. Similarly guess the cube roots of 4913, 12167, and 32768.
Method: Find the range from the number of digits, then use the units digit to identify the cube root's units digit.
∛1331: Between 10³=1000 and 20³=8000. Ends in 1 → root ends in 1. Only option: 11. Verify: 11³=1331 ✓
∛4913: Between 10³=1000 and 20³=8000. Ends in 3 → root ends in 7. Only option: 17. Verify: 17³=4913 ✓
∛12167: Between 20³=8000 and 30³=27000. Ends in 7 → root ends in 3. Only option: 23. Verify: 23³=12167 ✓
∛32768: Between 30³=27000 and 40³=64000. Ends in 8 → root ends in 2. Only option: 32. Verify: 32³=32768 ✓
∛1331: Between 10³=1000 and 20³=8000. Ends in 1 → root ends in 1. Only option: 11. Verify: 11³=1331 ✓
∛4913: Between 10³=1000 and 20³=8000. Ends in 3 → root ends in 7. Only option: 17. Verify: 17³=4913 ✓
∛12167: Between 20³=8000 and 30³=27000. Ends in 7 → root ends in 3. Only option: 23. Verify: 23³=12167 ✓
∛32768: Between 30³=27000 and 40³=64000. Ends in 8 → root ends in 2. Only option: 32. Verify: 32³=32768 ✓
Q5. Which is greatest? Explain your reasoning.
(i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²
(i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²
Using formulas (no full computation):
Squares: n² − (n−1)² = 2n − 1
(iii) 67² − 66² = 2×67 − 1 = 133
(iv) 43² − 42² = 2×43 − 1 = 85
Cubes: n³ − (n−1)³ = 3n² − 3n + 1
(i) 67³ − 66³ = 3×67² − 3×67 + 1 = 3×4489 − 201 + 1 = 13467 − 200 = 13268
(ii) 43³ − 42³ = 3×43² − 3×43 + 1 = 3×1849 − 129 + 1 = 5547 − 128 = 5420
Greatest: (i) 67³ − 66³ = 13268. Consecutive cube differences grow much faster than consecutive square differences.
Squares: n² − (n−1)² = 2n − 1
(iii) 67² − 66² = 2×67 − 1 = 133
(iv) 43² − 42² = 2×43 − 1 = 85
Cubes: n³ − (n−1)³ = 3n² − 3n + 1
(i) 67³ − 66³ = 3×67² − 3×67 + 1 = 3×4489 − 201 + 1 = 13467 − 200 = 13268
(ii) 43³ − 42³ = 3×43² − 3×43 + 1 = 3×1849 − 129 + 1 = 5547 − 128 = 5420
Greatest: (i) 67³ − 66³ = 13268. Consecutive cube differences grow much faster than consecutive square differences.
📋 Chapter 1 Summary — A Square and A Cube
- A square number (perfect square) is \(n^2 = n \times n\). The squares of natural numbers are 1, 4, 9, 16, 25, …
- Perfect squares end only in 0, 1, 4, 5, 6, or 9. If a number ends in 2, 3, 7, or 8 it is definitely not a perfect square. Squares can only end with an even number of zeros.
- The square root is the inverse of squaring: \(\sqrt{n^2} = n\). Every perfect square has two integer square roots (positive and negative). The symbol √ denotes the positive root.
- A number is a perfect square if its prime factors can be split into two identical groups.
- The sum of first n odd numbers = n². Consecutive square differences give consecutive odd numbers.
- A cube number (perfect cube) is \(n^3 = n \times n \times n\). Examples: 1, 8, 27, 64, 125, …
- A number is a perfect cube if its prime factors can be split into three identical groups.
- The cube root is denoted by \(\sqrt[3]{\ }\). For example, \(\sqrt[3]{27} = 3\). Each cube has a unique cube root (unlike squares, cube roots are unique even including negatives).
- The Hardy–Ramanujan number 1729 is the smallest number expressible as the sum of two positive cubes in two different ways: \(1^3 + 12^3 = 9^3 + 10^3\).
🎯 Square Pairs — Puzzle Activity
The numbers 3, 6, 10, 15, 1 are arranged so that each adjacent pair sums to a perfect square: 3+6=9, 6+10=16, 10+15=25, 15+1=16.
3
+
6
+
10
+
15
+
1
→ 9, 16, 25, 16 (all squares!)
Challenge 1: Arrange the numbers 1 to 17 in a row (no repeats) so every adjacent pair sums to a perfect square. Can you do it in more than one way?
One valid arrangement: 1–8–17–4–5–11–14–2–7–9–16–3–6–10–15–10–…
A known solution: 1, 8, 17, 4, 5, 11, 14, 2, 7, 9, 16, 3, 6, 10, 15, 1… (verify: 1+8=9✓, 8+17=25✓, 17+4=21✗). The arrangement must be found by trial and there is exactly one way (up to reversal) to arrange 1–15, and new solutions emerge as more numbers are included. Try it systematically!
Challenge 2: Arrange numbers 1 to 32 in a circle so every adjacent pair (including the wrap-around) sums to a perfect square.
Model this as a graph problem: create a node for each number 1–32, and connect two nodes with an edge if they sum to a perfect square. A circular arrangement exists if and only if this graph has a Hamiltonian cycle (a cycle that visits every node exactly once). This was proven to exist for 1–32!
📝
Competency-Based Questions
A school building has a basement storage room that is designed as a perfect cube. Its volume is 13824 cubic metres. On the wall, someone has painted the number 1729 with a note: "This is special — find out why!" A student, Kezia, claims: "I can find the side length of the room just by looking at the last digit of 13824, the digit count, and the cube root unit-digit trick — no calculator needed."
Q1. Using Kezia's method, what is the side length of the cubic room?
L3 Apply
Answer: (B) 24 m
13824 is a 5-digit number → cube root is between 20 (20³=8000) and 30 (30³=27000).
Last digit of 13824 is 4 → cube root ends in 4 → only 24 in range 20–30.
Verify: 24³ = 24×24×24 = 576×24 = 13824 ✓
13824 is a 5-digit number → cube root is between 20 (20³=8000) and 30 (30³=27000).
Last digit of 13824 is 4 → cube root ends in 4 → only 24 in range 20–30.
Verify: 24³ = 24×24×24 = 576×24 = 13824 ✓
Q2. Why is 1729 considered special? Express it as the sum of two cubes in two ways.
L1 Remember
1729 is the Hardy–Ramanujan number — the smallest positive integer expressible as the sum of two positive cubes in two different ways:
Way 1: 1³ + 12³ = 1 + 1728 = 1729
Way 2: 9³ + 10³ = 729 + 1000 = 1729
Way 1: 1³ + 12³ = 1 + 1728 = 1729
Way 2: 9³ + 10³ = 729 + 1000 = 1729
Q3. The room's floor area is 24² = 576 m². Use prime factorisation to verify 576 is a perfect square and find √576.
L2 Understand
576 = 2 × 288 = 2² × 144 = 2² × 4 × 36 = 2⁴ × 36 = 2⁴ × 4 × 9 = 2⁶ × 3² = (2³)² × 3² = (8×3)² = 24².
Split into two groups: (2³ × 3) × (2³ × 3) = 24 × 24.
√576 = 24 ✓ — consistent with the room's side length.
Split into two groups: (2³ × 3) × (2³ × 3) = 24 × 24.
√576 = 24 ✓ — consistent with the room's side length.
Q4. How many numbers lie between 24² and 25²? Between 24³ and 25³? Which gap is larger, and why?
L4 Analyse
Between n² and (n+1)² lie 2n numbers: between 576 and 625 → 48 numbers.
Between 24³=13824 and 25³=15625: gap = 15625−13824−1 = 1800 numbers.
The cube gap (25³−24³ = 3×25²−3×25+1 = 1876−75+1 = 1801) is far larger because cubes grow much faster than squares. This shows why cube root estimation must be more careful than square root estimation.
Between 24³=13824 and 25³=15625: gap = 15625−13824−1 = 1800 numbers.
The cube gap (25³−24³ = 3×25²−3×25+1 = 1876−75+1 = 1801) is far larger because cubes grow much faster than squares. This shows why cube root estimation must be more careful than square root estimation.
Q5 (HOT). The volume of the room (13824 m³) is both a perfect cube AND a perfect square. Verify this, then prove that any number of the form n⁶ is both a perfect square and a perfect cube.
L6 Create
13824 = 24³ = (24)³. Is it a perfect square? 13824 = 2⁶ × 3² × (wait: 24 = 2³×3, so 24³ = 2⁹×3³). Since 2⁹×3³ has odd exponents (9 and 3), it is not a perfect square. So the premise of this specific number being both is incorrect — but the general proof stands:
General Proof: For any number n⁶:
• Perfect square: n⁶ = (n³)² ✓
• Perfect cube: n⁶ = (n²)³ ✓
So any sixth power is simultaneously a perfect square and a perfect cube. Examples: 64 = 2⁶ = 8² = 4³; 729 = 3⁶ = 27² = 9³.
General Proof: For any number n⁶:
• Perfect square: n⁶ = (n³)² ✓
• Perfect cube: n⁶ = (n²)³ ✓
So any sixth power is simultaneously a perfect square and a perfect cube. Examples: 64 = 2⁶ = 8² = 4³; 729 = 3⁶ = 27² = 9³.
🔍 Assertion–Reason Questions
A: Both true; R explains A. B: Both true; R does NOT explain A. C: A true, R false. D: A false, R true.
Assertion: ∛1728 = 12.
Reason: 1728 = 2⁶ × 3³ = (2² × 3)³ = 12³, and the prime factors can be split into three identical groups each equalling 12.
Reason: 1728 = 2⁶ × 3³ = (2² × 3)³ = 12³, and the prime factors can be split into three identical groups each equalling 12.
Answer: A — Both true; R correctly justifies A using prime factorisation: 1728 = (2²×3)³ = 12³, so ∛1728 = 12.
Assertion: A cube cannot end with exactly two zeros (e.g., 100, 900 are not perfect cubes).
Reason: Trailing zeros in cubes must appear in multiples of 3, because (10k)³ = 1000k³ adds exactly 3 zeros.
Reason: Trailing zeros in cubes must appear in multiples of 3, because (10k)³ = 1000k³ adds exactly 3 zeros.
Answer: A — Both are true and R correctly explains A. Since 10³ = 1000 contributes 3 zeros, any trailing zeros from powers of 10 in a cube must come in groups of 3.
Assertion: 1729 is the smallest number expressible as the sum of two cubes in exactly one way.
Reason: Ramanujan observed that 1729 = 1³ + 12³ = 9³ + 10³.
Reason: Ramanujan observed that 1729 = 1³ + 12³ = 9³ + 10³.
Answer: D — The Assertion is false: 1729 is famous for being expressible as sum of two cubes in two ways (not one). The Reason is true and accurately states the two representations.
Frequently Asked Questions — Squares, Square Roots, Cubes and Cube Roots
What is Cubes, Cube Roots & Chapter Exercises in NCERT Class 8 Mathematics?
Cubes, Cube Roots & Chapter Exercises is a key concept covered in NCERT Class 8 Mathematics, Chapter 1: Squares, Square Roots, Cubes and Cube Roots. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Cubes, Cube Roots & Chapter Exercises step by step?
To solve problems on Cubes, Cube Roots & Chapter Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 1: Squares, Square Roots, Cubes and Cube Roots?
The essential formulas of Chapter 1 (Squares, Square Roots, Cubes and Cube Roots) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Cubes, Cube Roots & Chapter Exercises important for the Class 8 board exam?
Cubes, Cube Roots & Chapter Exercises is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Cubes, Cube Roots & Chapter Exercises?
Common mistakes in Cubes, Cube Roots & Chapter Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Cubes, Cube Roots & Chapter Exercises?
End-of-chapter NCERT exercises for Cubes, Cube Roots & Chapter Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 1, and solve at least one previous-year board paper to consolidate your understanding.
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