Chapter Overview
1729
Hardy–Ramanujan Number (sum of two cubes)
1–20
Cubes table (1³ to 20³)
EX
Examples: perfect cubes & prime factor method
2
Main methods: prime factorisation & pattern recognition
What you'll learn
Definition & Patterns
What are cubes and cube roots; patterns in cube endings and sums of odd numbers giving cubes.
Prime-factor method
How to test perfect cubes and find cube roots via grouping prime factors in triples.
Smallest multiplier/divisor
How to find the least number to multiply/divide to get a perfect cube (by adjusting prime exponents to multiples of 3).
Applications
Volume-side relationships of cubes and useful numeric identities (e.g., sums of consecutive odd numbers).
Quick context
This chapter introduces cubes (numbers of form \(n^3\)) and cube roots (the inverse operation). You will see examples, interesting numerical patterns (like Hardy–Ramanujan type facts), and techniques to decide if a number is a perfect cube and to compute cube roots using prime factorisation. Practical tasks include finding the least multiplier or divisor to make a number a cube and interpreting relationship between volume and side of cubes.
Comprehensive Chapter Summary
1. Introduction to cubes
Cube of a number \(n\) is \(n^3 = n \times n \times n\). Perfect cubes among small numbers: \(1, 8, 27, 64, 125, 216, 343, 512, 729, 1000,\dots\). The chapter highlights interesting numbers like 1729 — the smallest number expressible as sum of two cubes in two different ways (\(1729 = 1^3+12^3 = 9^3+10^3\)). :contentReference[oaicite:0]{index=0}
2. Recognising perfect cubes
Prime factor rule
A number is a perfect cube if, in its prime factorisation, every prime exponent is a multiple of 3. Example: \(216=2^3\cdot3^3\) so \(216=(2\cdot3)^3=6^3\). :contentReference[oaicite:1]{index=1}
Unit-digit patterns
The one's digit of cubes follows patterns (e.g., numbers ending in 2 have cubes ending in 8). Try examples like \(3331^3\), \(1024^3\). :contentReference[oaicite:2]{index=2}
3. Smallest multiplier / divisor to get a cube
To make a number \(N\) a perfect cube, adjust its prime exponents so each becomes a multiple of 3. If a prime \(p\) appears \(e\) times and \(e \bmod 3 = r\), you multiply by \(p^{(3-r)}\). Similarly, to divide to get a cube, divide out the primes whose exponents cannot be grouped in threes. Worked examples: 392 requires multiplication by 7 to become \(2744=14^3\); 1188 must be divided by 44 to yield 27.
4. Cube roots
Cube root is denoted \(\sqrt[3]{\cdot}\). Using prime factorisation, group primes in triples and multiply the base primes left after grouping. Examples: \(\sqrt[3]{3375}=15\) since \(3375=3^3\cdot5^3\). Other examples: \(\sqrt[3]{8000}=20,\ \sqrt[3]{13824}=24.\)
5. Patterns & identities
Some numeric patterns: sums of consecutive odd numbers give cubes (e.g., \(1=1^3,\;3+5=2^3,\;7+9+11=3^3\)). Differences of cubes follow a factor pattern: \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Explore ending-digit behaviour and growth comparisons like \(m^2 < m^3\) for suitable integers. :contentReference[oaicite:5]{index=5}
6. What we practiced
- Identify perfect cubes, compute cube roots by prime factor method.
- Find least numbers to multiply/divide to obtain perfect cubes.
- Understand patterns: Hardy–Ramanujan number, unit-digit patterns, sums of odd numbers producing cubes.
Questions and Answers — Chapter 6
Short Questions (brief answers)
Q1. What is a perfect cube? Give two examples.
Answer: A perfect cube is \(n^3\) for some integer \(n\). Examples: \(27 = 3^3,\;512 = 8^3\).
Q2. Is 500 a perfect cube?
Answer: No. Prime factors: \(500=2^2\cdot5^3\) (2's exponent not multiple of 3). :contentReference[oaicite:6]{index=6}
Q3. Find \(\sqrt[3]{8000}\).
Answer: \(20\). (Because \(8000=2^6\cdot5^3=(2^2\cdot5)^3=(4\cdot5)^3=20^3\)). :contentReference[oaicite:7]{index=7}
Q4. What is the cube of 12?
Answer: \(12^3 = 1728\).
Q5. Give the cube root of 3375.
Answer: \(15\), since \(3375=3^3\cdot5^3\). :contentReference[oaicite:8]{index=8}
Q6. Is 243 a perfect cube?
Answer: No; \(243=3^5\) — after grouping triples a \(3^2\) remains. :contentReference[oaicite:9]{index=9}
Q7. State the prime-factor test for perfect cubes.
Answer: In prime factorisation, every exponent must be a multiple of 3. :contentReference[oaicite:10]{index=10}
Q8. What is 20³?
Answer: \(8000\).
Q9. What is the cube root of 13824?
Answer: \(24\). (See prime factor method). :contentReference[oaicite:11]{index=11}
Q10. If a number ends with digit 8, can it be a cube? (Give example or counter)
Answer: Yes — for example \(2^3=8\) ends with 8; but many numbers ending in 8 are not necessarily cubes. See unit-digit patterns. :contentReference[oaicite:12]{index=12}
Q11. Is 27000 a perfect cube?
Answer: Yes, \(27000 = 30^3\). (From Exercise list). :contentReference[oaicite:13]{index=13}
Q12. What is special about 1729?
Answer: It's the smallest number expressible as sum of two cubes in two different ways: \(1729=1^3+12^3=9^3+10^3\). :contentReference[oaicite:14]{index=14}
Q13. How many cubes are there between 1 and 1000?
Answer: Ten cubes (1³ to 10³). :contentReference[oaicite:15]{index=15}
Q14. Is 10648 a perfect cube? If yes, what is its cube root?
Answer: Yes; \(\sqrt[3]{10648}=22\). (Exercise lists 10648 as a cube). :contentReference[oaicite:16]{index=16}
Q15. What is the cube of 5?
Answer: \(5^3 = 125\).
Medium Questions — (Answer in ~3 marks)
Q1. Using prime factorisation find the cube root of 46656.
Answer: \(46656=2^6\cdot3^6=(2^2\cdot3^2)^3=(4\cdot9)^3=36^3\). So \(\sqrt[3]{46656}=36\). :contentReference[oaicite:17]{index=17}
Q2. Find the smallest number by which 392 must be multiplied to get a perfect cube.
Answer: \(392=2^3\cdot7^2\). To make exponents multiples of 3, multiply by \(7\). So multiply by 7 to get \(2744=14^3\). :contentReference[oaicite:18]{index=18}
Q3. By which smallest number should 53240 be divided to obtain a perfect cube?
Answer: \(53240=2^3\cdot11^3\cdot5\). Divide by 5 to remove singleton 5 → quotient \(10648=22^3\). So divide by 5. :contentReference[oaicite:19]{index=19}
Q4. Find cube root of 110592 by prime factorisation (hint: used in exercises).
Answer: \(110592 = 2^{12}\cdot3^3 = (2^4\cdot3)^3 = (16\cdot3)^3 = 48^3\). So cube root = 48. (Grouping exponents by 3). :contentReference[oaicite:20]{index=20}
Q5. Find the smallest number by which 68600 must be multiplied to get a perfect cube.
Answer: \(68600=2^3\cdot5^2\cdot7^3\). Missing one 5 to make exponent 3 → multiply by 5. Then product \(343000=70^3\). :contentReference[oaicite:21]{index=21}
Q6. Check whether 1000 and 9000 are perfect cubes.
Answer: \(1000=10^3\) is a perfect cube. \(9000=2^3\cdot3^2\cdot5^3\) — exponent of 3 is 2, so NOT a perfect cube. (See exercise lists). :contentReference[oaicite:22]{index=22}
Q7. Find \(\sqrt[3]{175616}\) by prime factorisation.
Answer: \(175616 = 2^9\cdot7^3 = (2^3\cdot7)^3=(8\cdot7)^3=56^3\). So cube root = 56. :contentReference[oaicite:23]{index=23}
Q8. Express 63 as sum of consecutive odd numbers using the cube-sum pattern.
Answer: \(63 = 3^3 + \text{(pattern)}\) — use the pattern of sums of odd numbers: \(21+23+19 = ?\) (Students write the sequence that sums to 63: \(11+13+15+... \) — standard exercise from chapter). :contentReference[oaicite:24]{index=24}
Q9. Decide true/false: "Cube of any odd number is even."
Answer: False. Cube of odd is odd (odd×odd×odd remains odd). :contentReference[oaicite:25]{index=25}
Q10. Find the cube root of 46656 using prime factorisation (exercise item).
Answer: See medium Q1: cube root = 36. (Worked earlier). :contentReference[oaicite:26]{index=26}
Q11. If \(N = 2^4\cdot3^2\cdot5\), what smallest multiplier makes \(N\) a perfect cube?
Answer: Exponents mod 3 are: 2 (for 2), 2 (for 3), 1 (for 5). Need multiply by \(2^2\cdot3^1\cdot5^2 = 4\cdot3\cdot25 = 300\). So multiply by 300. (Generic method.).
Q12. Why is 243 not a cube though it is a high power of 3?
Answer: \(243=3^5\). Since exponent 5 is not a multiple of 3, it's not a perfect cube. :contentReference[oaicite:27]{index=27}
Q13. Find the cube root of 91125 from the exercise list.
Answer: \(91125 = 3^3\cdot5^6 = (3\cdot5^2)^3 = (3\cdot25)^3 = 75^3\). So cube root = 75. :contentReference[oaicite:28]{index=28}
Q14. Show that 343000 is a perfect cube and find its cube root.
Answer: From example: \(343000 = 2^3\cdot5^3\cdot7^3 = (2\cdot5\cdot7)^3 = 70^3\). Therefore cube root = 70. :contentReference[oaicite:29]{index=29}
Q15. For \(n\in\mathbb{N}\), check whether \(n^2 < n^3\) always holds. Explain.
Answer: For \(n\ge2\), \(n^3 = n^2\cdot n > n^2\). For \(n=0,1\) equality/inequality differs: \(0^2=0^3\), \(1^2=1^3\). So statement true for \(n\ge2\). :contentReference[oaicite:30]{index=30}
Long Questions — (Detailed answers)
Q1. Explain in detail how to use prime factorisation to test whether a number is a perfect cube and find the smallest multiplier to make it a cube. Illustrate with 68600.
Answer:
Step 1: Prime-factorise: \(68600 = 2^3\cdot5^2\cdot7^3\). Step 2: Check exponents: for a perfect cube each exponent must be multiple of 3. Here exponents are 3 (OK), 2 (not OK), 3 (OK). Step 3: For exponent 2, need \(3-2=1\) extra factor of 5. So multiply by \(5\) to get \(68600\times5=343000\). After multiplication prime exponents become \(2^3\cdot5^3\cdot7^3\), all multiples of 3. Hence \(343000=(2\cdot5\cdot7)^3=70^3\). So the smallest multiplier is 5 and cube root of resulting cube is 70. :contentReference[oaicite:31]{index=31}
Q2. Work out fully: find the smallest number by which 1188 must be divided so that quotient is a perfect cube and state the cube root of the quotient.
Answer:
Factorise: \(1188=2^2\cdot3^3\cdot11\). Exponents: 2 (for 2), 3 (for 3), 1 (for 11). To make exponents multiples of 3 by division, remove the primes with exponents not divisible by 3 — remove \(2^2\) and \(11^1\) i.e. divide by \(2^2\cdot11 = 44\). Then quotient = \(1188/44 = 27 = 3^3\). Cube root = 3. So smallest divisor = 44. :contentReference[oaicite:32]{index=32}
Q3. Prove the formula \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and give an example with numbers.
Answer:
Expand RHS: \((a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3\). Example: let \(a=5, b=2\): LHS \(=125-8=117\). RHS \((5-2)(25+10+4)=3\times39=117\). Identity holds.
Q4. Using the cube-sum pattern (sum of odd numbers), show how 125 can be obtained as a sum of consecutive odd numbers.
Answer:
125 = \(5^3\). The pattern says sum of first \(n\) odd numbers equals \(n^2\), and certain consecutive-odd sums produce cubes: \(21+23+25+27+29=125\). Indeed, these five consecutive odd numbers sum to 125, exhibiting the pattern that the sum of \(n\) consecutive odd numbers (appropriately chosen) can give \(n^3\) for \(n=5\). (Chapter activities). :contentReference[oaicite:33]{index=33}
Q5. Demonstrate with full working how \(\sqrt[3]{74088}=42\) using prime factorisation.
Answer:
Prime factorise \(74088 = 2^3\cdot3^3\cdot7^3\). Group each prime exponent into triples: \((2^3)(3^3)(7^3) = (2\cdot3\cdot7)^3\). Therefore \(\sqrt[3]{74088} = 2\cdot3\cdot7 = 42.\) :contentReference[oaicite:34]{index=34}
Q6. A cuboid has dimensions \(15\times30\times15\) cm. How many such cuboids are needed to make a perfect cube? Explain.
Answer:
Volume of each cuboid \(=15\times30\times15=6750 = 2\times3^3\times5^3\) (prime factorisation gives one 2 and three 3s and three 5s). To make the combined volume a perfect cube, the exponent of 2 must be a multiple of 3. Currently exponent of 2 is 1; multiplying by 4 (i.e., using 4 such cuboids) provides \(2^3\). Thus 4 cuboids are required to make a cube. (See worked example). :contentReference[oaicite:35]{index=35}
Q7. Show with steps how to find \(\sqrt[3]{13824}\).
Answer:
\(13824 = 2^9\cdot3^3\) (since \(13824 = 2^9\times3^3\)). Grouping gives \((2^3\cdot3)^3 = (8\cdot3)^3 = 24^3\). So cube root = 24. :contentReference[oaicite:36]{index=36}
Q8. From the chapter: list ten cubes from \(1^3\) to \(10^3\) and explain why only ten cubes lie below 1000.
Answer:
1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000. Below 1000, only cubes of numbers 1–9 are strictly less than 1000; 10³ equals 1000. So there are 10 cubes from 1 to 1000 inclusive (1³ through 10³). :contentReference[oaicite:37]{index=37}
Q9. How do unit-digit patterns help to quickly reject non-cubes? Illustrate with 243 and 500.
Answer:
Study of unit digits of cubes shows allowed endings (e.g. 0,1,8,7,4,5,6,3,2,9 based on base digit). But prime exponent check is faster: 243 = \(3^5\) (not cube), 500 = \(2^2\cdot5^3\) (2's exponent not multiple of 3) → both not cubes. Unit digits alone are heuristic; prime-factor test is conclusive. :contentReference[oaicite:38]{index=38}
Q10. Explain the concept of Hardy–Ramanujan numbers and give one example besides 1729.
Answer:
Hardy–Ramanujan numbers are integers expressible as sum of two positive cubes in more than one way. 1729 is the smallest. Another example is 4104 which equals \(2^3+16^3=9^3+15^3\). These are interesting because they show different representations as sums of cubes. :contentReference[oaicite:39]{index=39}
Q11. Given \(N=2^5\cdot3^4\cdot7\), find the smallest number to multiply with \(N\) to get a perfect cube and find cube root of the result.
Answer:
Exponents modulo 3: 5→2 so need 1 more 2; 4→1 so need 2 more 3s; 1→1 so need 2 more 7s. Multiply by \(2^1\cdot3^2\cdot7^2 = 2\cdot9\cdot49=882\). New number is cube of \(2^{(5+1)/3}3^{(4+2)/3}7^{(1+2)/3} = 2^2\cdot3^2\cdot7^1 = 4\cdot9\cdot7 = 252\). So cube root = 252. (Method illustrated).
Q12. Using examples from the chapter, explain how to find the least number to divide \(53240\) to get a perfect cube.
Answer:
Prime factor \(53240 = 2^3\cdot11^3\cdot5^1\). Exponent of 5 is 1; dividing by 5 removes the lone 5, leaving all exponents multiples of 3. Thus divide by 5 → quotient \(10648=22^3\). So smallest divisor = 5. :contentReference[oaicite:40]{index=40}
Q13. Evaluate and explain: Does any perfect cube end with two zeros? (Reference tasks in the chapter.)
Answer:
If a number ends with two zeros, it's divisible by 100 = \(2^2\cdot5^2\). For it to be a cube, exponents of 2 and 5 in prime factorisation must be multiples of 3. Since 2 and 5 each contribute at least exponent 2, we need extra factors to make them multiples of 3 (e.g., exponent 3 is possible if multiplied suitably). So while some numbers with two ending zeros can be cubes (e.g., 1000 ends with three zeros and is \(10^3\)), a cube ending with exactly two zeros is not possible because exponent-of-5 and 2 would not both be multiples of 3 with exactly two trailing zeros. (Chapter has related true/false items). :contentReference[oaicite:41]{index=41}
Q14. Prove that if each prime factor of a number appears three times, the number is a perfect cube. Give an example.
Answer:
If \(N=\prod p_i^{3k_i}\) then \(N=\big(\prod p_i^{k_i}\big)^3\), hence a perfect cube. Example: \(216=2^3\cdot3^3=(2\cdot3)^3=6^3\). (Seen in chapter examples). :contentReference[oaicite:42]{index=42}
Q15. Discuss patterns observed in cubes of numbers ending with digits 1–9; illustrate how the one's digit of cube depends on unit digit of original number.
Answer:
Compute single-digit cubes: \(0^3=0,1^3=1,2^3=8,3^3=27 (units 7),4^3=64 (units 4),5^3=125 (units 5),6^3=216 (units 6),7^3=343 (units 3),8^3=512 (units 2),9^3=729 (units 9)\). The unit digit of the cube is determined only by unit digit of the original number. This helps quick checks (chapter 'Try these' tasks). :contentReference[oaicite:43]{index=43}
Source (chapter used to prepare the above questions & answers):
