Cubes and Cube Roots — Questions and Answers
Welcome to HSLC Guru. This lesson gives the full, step-by-step solutions to ASSEB (Assam State School Education Board) Class 8 General Mathematics Chapter 7, Cubes and Cube Roots — every worked example, every “Try Yourself” activity, and all of Exercise 7.1 and Exercise 7.2.
Summary
The number obtained when a number is multiplied by itself three times is called the cube of that number; for example $2^3 = 2 \times 2 \times 2 = 8$ and $3^3 = 27$. Numbers such as 1, 8, 27, 64, 125 are called perfect cubes. There are only 10 perfect cubes from 1 to 1000.
To check whether a number is a perfect cube, write its prime factorisation and group the prime factors in triples. If every prime factor forms a complete triple, the number is a perfect cube. When it is not, the same method tells us the smallest number by which we must multiply or divide to make it a perfect cube.
The inverse operation of cubing is finding the cube root, written with the symbol $\sqrt[3]{\ }$. A cube root can be found either by the prime factorisation method or by the method of estimation. The chapter also covers that cubes of even numbers are even and cubes of odd numbers are odd, some amazing cube-square patterns, and the famous Ramanujan number 1729.
Summary: This ASSEB Class 8 General Mathematics Chapter 7 (Cubes and Cube Roots) solution explains cube numbers, testing perfect cubes by prime factorisation, finding the smallest number to multiply or divide by, cube roots by the prime factorisation and estimation methods, cube-square patterns and the Ramanujan number 1729, with full worked answers to Exercise 7.1 and Exercise 7.2.
Textbook Questions and Answers
7.1 Cube Numbers — Basic Idea
How many perfect cubes are there from 1 to 1000? Complete the table.
Answer: Cubing the numbers 1 to 10 gives $1^3=1$, $2^3=8$, $3^3=27$, $4^3=64$, $5^3=125$, $6^3=216$, $7^3=343$, $8^3=512$, $9^3=729$, $10^3=1000$. So there are only 10 perfect cubes from 1 to 1000.
| Number | Cube | Cube Number |
|---|---|---|
| 1 | $1^3$ | 1 |
| 2 | $2^3$ | 8 |
| 3 | $3^3$ | 27 |
| 4 | $4^3$ | 64 |
| 5 | $5^3$ | 125 |
| 6 | $6^3$ | 216 |
| 7 | $7^3$ | 343 |
| 8 | $8^3$ | 512 |
| 9 | $9^3$ | 729 |
| 10 | $10^3$ | 1000 |
Is 50 a cube number? Is 1728 a cube number?
Answer: $50 = 2 \times 5 \times 5$; the same prime factor does not occur three times, so 50 is not a cube number. But $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 2^3 \times 3^3 = (2 \times 2 \times 3)^3 = 12^3$, so 1728 = 12³ is a perfect cube — it is the cube of 12.
Solved Examples (Examples 1–6)
Example 1: Examine whether 100 is a cube number or not.
Answer: $100 = 10 \times 10 = 2 \times 5 \times 2 \times 5 = 2 \times 2 \times 5 \times 5$. Here 2 and 5 each appear only twice, so they cannot be grouped in triples; hence 100 is not a cube number. But since $100 = 10^2$, it is a square number. (Note $64 = 4^3 = 8^2$, so 64 is both a cube and a square.)
Example 2: Is 243 a perfect cube? If not, find the smallest natural number by which 243 must be multiplied to get a perfect cube.
Answer: $243 = 3 \times 3 \times 3 \times 3 \times 3 = 3^5$. Grouping in triples gives one complete triple ($3^3$) but $3 \times 3$ is left over, so 243 is not a perfect cube. One more 3 is needed to complete a triple, so it must be multiplied by the smallest number 3; then $243 \times 3 = 729 = 9^3$.
Example 3: Is 2187 a perfect cube? If not, find the smallest natural number by which 2187 must be divided to get a perfect cube.
Answer: $2187 = 3^7$. Two triples ($3^3 \times 3^3 = 3^6$) are formed and one 3 is left over, so 2187 is not a perfect cube. Removing one 3 gives $3^6 = 729 = 9^3$, which is a perfect cube; so it must be divided by the smallest number 3.
Example 4: Determine the smallest number by which 35000 must be divided to make it a perfect cube.
Answer: $35000 = 35 \times 1000 = 5 \times 7 \times 10 \times 10 \times 10 = 5 \times 7 \times 2^3 \times 5^3$. Here $2^3$ and $5^3$ are complete triples, but one extra 5 and one 7 remain (each occurs once outside the triples). So it must be divided by $5 \times 7 = 35$; then $35000 \div 35 = 1000 = 10^3$.
Example 5: Find the cube root of 15625.
Answer: $15625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^3 \times 5^3 = (5 \times 5)^3 = 25^3$. So $\sqrt[3]{15625} = 25$.
Example 6: Find the cube root of 12167 by the method of estimation.
Answer: Making groups of three from the right gives $\overline{12}\ \overline{167}$. The unit digit of the first group is 7; a number whose cube ends in 7 has cube root ending in 3 ($3^3 = 27$), so the unit digit is 3. For the left group 12, the largest cube not exceeding 12 is 8 ($=2^3$), whose cube root is 2, so the tens digit is 2. Hence $\sqrt[3]{12167} = 23$ (check: $23^3 = 12167$).
Additional prime-factorisation example: Find $\sqrt[3]{13824}$.
Answer: $13824 = 2^9 \times 3^3 = 2^3 \times 2^3 \times 2^3 \times 3^3 = (2 \times 2 \times 2 \times 3)^3 = 24^3$; so $\sqrt[3]{13824} = 24$.
7.3 Some Amazing Patterns (Activity)
Write the relation between cube numbers and square numbers.
Answer: The sum of the cubes of the first few natural numbers equals the square of their sum —
$$1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$$
- $1^3 = 1 = 1^2$
- $1^3 + 2^3 = 9 = (1+2)^2 = 3^2$
- $1^3 + 2^3 + 3^3 = 36 = (1+2+3)^2 = 6^2$
- $1^3 + 2^3 + 3^3 + 4^3 = 100 = (1+2+3+4)^2 = 10^2$
- $1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225 = 15^2$
Express cube numbers as the sum of consecutive odd numbers.
Answer: Every cube number can be written as a sum of consecutive odd numbers —
- $1 = 1^3$
- $3 + 5 = 8 = 2^3$
- $7 + 9 + 11 = 27 = 3^3$
- $13 + 15 + 17 + 19 = 64 = 4^3$
- $21 + 23 + 25 + 27 + 29 = 125 = 5^3$
- $31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3$
What is special about the Ramanujan number 1729?
Answer: 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways — $1729 = 12^3 + 1^3 = 1728 + 1$ and $1729 = 10^3 + 9^3 = 1000 + 729$. It is called the Ramanujan number. Similarly, $4104 = 16^3 + 2^3 = 15^3 + 9^3$.
Try Yourself
Find the cube roots by the prime factorisation method — (a) 512 (b) 27000 (c) 110592 (d) 46656 (e) 175616
Answer:
- (a) $512 = 2^9 = (2^3)^3 = 8^3$, so $\sqrt[3]{512} = 8$.
- (b) $27000 = 2^3 \times 3^3 \times 5^3 = (2 \times 3 \times 5)^3 = 30^3$, so $\sqrt[3]{27000} = 30$.
- (c) $110592 = 2^{12} \times 3^3 = (2^4 \times 3)^3 = 48^3$, so $\sqrt[3]{110592} = 48$.
- (d) $46656 = 2^6 \times 3^6 = (2^2 \times 3^2)^3 = 36^3$, so $\sqrt[3]{46656} = 36$.
- (e) $175616 = 2^9 \times 7^3 = (2^3 \times 7)^3 = 56^3$, so $\sqrt[3]{175616} = 56$.
Find the cube roots by the method of estimation — (a) 4096 (b) 9261 (c) 13824 (d) 15625
Answer:
- (a) $\overline{4}\ \overline{096}$: unit 6 → cube root unit 6; largest cube $\le 4$ is 1 ($=1^3$) → tens 1; so $\sqrt[3]{4096} = 16$.
- (b) $\overline{9}\ \overline{261}$: unit 1 → unit 1; largest cube $\le 9$ is 8 ($=2^3$) → tens 2; so $\sqrt[3]{9261} = 21$.
- (c) $\overline{13}\ \overline{824}$: unit 4 → unit 4; largest cube $\le 13$ is 8 ($=2^3$) → tens 2; so $\sqrt[3]{13824} = 24$.
- (d) $\overline{15}\ \overline{625}$: unit 5 → unit 5; largest cube $\le 15$ is 8 ($=2^3$) → tens 2; so $\sqrt[3]{15625} = 25$.
Exercise 7.1
1. Which of the following numbers are perfect cubes?
(a) 500 (b) 1331 (c) 2025 (d) 6859 (e) 2376 (f) 8000
Answer: Check whether the prime factors group into triples —
- (a) $500 = 2^2 \times 5^3$ — 2 occurs only twice → not a perfect cube.
- (b) $1331 = 11^3$ → perfect cube ($\sqrt[3]{1331}=11$).
- (c) $2025 = 3^4 \times 5^2$ → not a perfect cube.
- (d) $6859 = 19^3$ → perfect cube ($\sqrt[3]{6859}=19$).
- (e) $2376 = 2^3 \times 3^3 \times 11$ — 11 occurs once → not a perfect cube.
- (f) $8000 = 2^6 \times 5^3 = 20^3$ → perfect cube ($\sqrt[3]{8000}=20$).
So the perfect cubes are (b) 1331, (d) 6859 and (f) 8000.
2. Find the smallest number by which each of the following must be multiplied to obtain a perfect cube.
(a) 675 (b) 256 (c) 100 (d) 72
Answer:
- (a) $675 = 3^3 \times 5^2$; one more 5 is needed to complete $5^3$ → multiply by 5; $675 \times 5 = 3375 = 15^3$.
- (b) $256 = 2^8$; one more 2 is needed to make $2^9$ → multiply by 2; $256 \times 2 = 512 = 8^3$.
- (c) $100 = 2^2 \times 5^2$; one more 2 and one more 5 are needed → multiply by $2 \times 5 = $ 10; $100 \times 10 = 1000 = 10^3$.
- (d) $72 = 2^3 \times 3^2$; one more 3 is needed → multiply by 3; $72 \times 3 = 216 = 6^3$.
3. Find the smallest number by which each of the following must be divided to obtain a perfect cube.
(a) 2401 (b) 8192 (c) 6561 (d) 1,00,000
Answer:
- (a) $2401 = 7^4 = 7^3 \times 7$; remove the extra 7 → divide by 7; $2401 \div 7 = 343 = 7^3$.
- (b) $8192 = 2^{13} = 2^{12} \times 2$; remove the extra 2 → divide by 2; $8192 \div 2 = 4096 = 16^3$.
- (c) $6561 = 3^8 = 3^6 \times 3^2$; remove the extra $3^2$ → divide by 9; $6561 \div 9 = 729 = 9^3$.
- (d) $100000 = 2^5 \times 5^5 = 2^3 \times 2^2 \times 5^3 \times 5^2$; remove $2^2 \times 5^2 = 100$ → divide by 100; $100000 \div 100 = 1000 = 10^3$.
4. Find the smallest number by which each of the following must be multiplied or divided to obtain a cube number.
(a) 250 (b) 675 (c) 1372 (d) 3000 (e) 153664
Answer: Consider both operations and take the smaller number —
- (a) $250 = 2 \times 5^3$; multiply by $2^2 = 4$, or divide by 2. Smaller is 2, so divide by 2; $250 \div 2 = 125 = 5^3$.
- (b) $675 = 3^3 \times 5^2$; multiply by 5 ($=3375=15^3$), or divide by $5^2 = 25$. Smaller is 5, so multiply by 5; $675 \times 5 = 3375 = 15^3$.
- (c) $1372 = 2^2 \times 7^3$; multiply by 2 ($=2744=14^3$), or divide by $2^2 = 4$. Smaller is 2, so multiply by 2; $1372 \times 2 = 2744 = 14^3$.
- (d) $3000 = 2^3 \times 3 \times 5^3$; multiply by $3^2 = 9$ ($=27000=30^3$), or divide by 3. Smaller is 3, so divide by 3; $3000 \div 3 = 1000 = 10^3$.
- (e) $153664 = 2^6 \times 7^4$; multiply by $7^2 = 49$, or divide by 7. Smaller is 7, so divide by 7; $153664 \div 7 = 21952 = 28^3$.
Exercise 7.2
1. Choose the correct option for each of the following —
(i) The unit place of the cube of 23 — (a) 3 (b) 6 (c) 7 (d) 9
Answer: (c) 7. The unit digit 3 has cube $3^3 = 27$, whose unit digit is 7 (and $23^3 = 12167$).
(ii) Which of the following numbers is a perfect cube? — (a) 243 (b) 216 (c) 392 (d) 8640
Answer: (b) 216, since $216 = 2^3 \times 3^3 = 6^3$. ($243 = 3^5$, $392 = 2^3 \times 7^2$ and 8640 are not perfect cubes.)
(iii) Which of the following numbers is not a perfect cube? — (a) 216 (b) 567 (c) 125 (d) 343
Answer: (b) 567. $567 = 3^4 \times 7$ is not a perfect cube. ($216 = 6^3$, $125 = 5^3$, $343 = 7^3$ are perfect cubes.)
(iv) The value of $\sqrt[3]{1000}$ — (a) 1 (b) 10 (c) 100 (d) 1000
Answer: (b) 10, since $1000 = 10^3$, so $\sqrt[3]{1000} = 10$.
(v) The value of $\sqrt[3]{27} + \sqrt[3]{64} + \sqrt[3]{125}$ — (a) 10 (b) 11 (c) 12 (d) 13
Answer: (c) 12. $\sqrt[3]{27} + \sqrt[3]{64} + \sqrt[3]{125} = 3 + 4 + 5 = 12$.
2. Find the cube roots of the following by the prime factorisation method —
(i) 125 (ii) 343 (iii) 2744 (iv) 10648 (v) 4096 (vi) 35937 (vii) 216000 (viii) 9261 (ix) 21952 (x) 6859
Answer:
- (i) $125 = 5^3$ → $\sqrt[3]{125} = 5$.
- (ii) $343 = 7^3$ → $\sqrt[3]{343} = 7$.
- (iii) $2744 = 2^3 \times 7^3 = (2 \times 7)^3 = 14^3$ → $\sqrt[3]{2744} = 14$.
- (iv) $10648 = 2^3 \times 11^3 = (2 \times 11)^3 = 22^3$ → $\sqrt[3]{10648} = 22$.
- (v) $4096 = 2^{12} = (2^4)^3 = 16^3$ → $\sqrt[3]{4096} = 16$.
- (vi) $35937 = 3^3 \times 11^3 = (3 \times 11)^3 = 33^3$ → $\sqrt[3]{35937} = 33$.
- (vii) $216000 = 2^6 \times 3^3 \times 5^3 = (2^2 \times 3 \times 5)^3 = 60^3$ → $\sqrt[3]{216000} = 60$.
- (viii) $9261 = 3^3 \times 7^3 = (3 \times 7)^3 = 21^3$ → $\sqrt[3]{9261} = 21$.
- (ix) $21952 = 2^6 \times 7^3 = (2^2 \times 7)^3 = 28^3$ → $\sqrt[3]{21952} = 28$.
- (x) $6859 = 19^3$ → $\sqrt[3]{6859} = 19$.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The value of $6^3$ — (a) 18 (b) 36 (c) 72 (d) 216
Answer: (d) 216.
2. Which of the following is a perfect cube? — (a) 100 (b) 125 (c) 150 (d) 200
Answer: (b) 125 ($= 5^3$).
3. The value of $\sqrt[3]{729}$ — (a) 7 (b) 8 (c) 9 (d) 27
Answer: (c) 9 ($729 = 9^3$).
4. The value of $\sqrt[3]{1728}$ — (a) 10 (b) 12 (c) 14 (d) 16
Answer: (b) 12 ($1728 = 12^3$).
5. The cube of an even number is — (a) an even number (b) an odd number (c) always zero (d) a negative number
Answer: (a) an even number.
6. If the unit place of a number is 4, the unit place of its cube is — (a) 2 (b) 4 (c) 6 (d) 8
Answer: (b) 4 ($4^3 = 64$).
7. The total number of perfect cubes from 1 to 1000 — (a) 8 (b) 9 (c) 10 (d) 11
Answer: (c) 10.
8. If the unit place of a number is 8, the unit place of its cube is — (a) 2 (b) 4 (c) 6 (d) 8
Answer: (a) 2 ($8^3 = 512$).
9. The Ramanujan number is — (a) 1000 (b) 1728 (c) 1729 (d) 4104
Answer: (c) 1729.
10. Which number is both a cube and a square? — (a) 16 (b) 36 (c) 64 (d) 81
Answer: (c) 64 ($64 = 4^3 = 8^2$).
Fill in the Blanks
- A number multiplied by itself ______ times gives its cube. — three
- $\sqrt[3]{343} = $ ______ — 7
- The cube of an odd number is an ______ number. — odd
- $5^3 = $ ______ — 125
- $1729 = 12^3 + $ ______$^3 = 10^3 + 9^3$. — 1
True or False
- 64 is both a cube number and a square number. — True
- 100 is a perfect cube. — False (it is a square number)
- The cube of an even number is odd. — False
- $\sqrt[3]{27} + \sqrt[3]{64} = 7$. — True ($3 + 4 = 7$)
- 2187 is a perfect cube. — False ($2187 = 3^7$)
Short Answer Questions
1. What is a cube number? Give an example.
Answer: The number obtained by multiplying a number by itself three times is called a cube number or a perfect cube; for example $3 \times 3 \times 3 = 27 = 3^3$, so 27 is a cube number.
2. Show by prime factorisation that 216 is a perfect cube.
Answer: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3 = (2 \times 3)^3 = 6^3$. Since every prime factor forms a triple, 216 is a perfect cube and $\sqrt[3]{216} = 6$.
3. What is the value of $1^3 + 2^3 + 3^3 + 4^3$, and which square does it equal?
Answer: $1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100 = (1+2+3+4)^2 = 10^2$. So it equals the square of 10.
4. Find $\sqrt[3]{4913}$ by the method of estimation.
Answer: $\overline{4}\ \overline{913}$: the unit digit of the first group is 3 → cube root unit digit 7 ($7^3 = 343$). For the left group 4, the largest cube not exceeding 4 is 1 ($=1^3$) → tens digit 1. Hence $\sqrt[3]{4913} = 17$ (check: $17^3 = 4913$).
Key Terms
| Term | Meaning |
|---|---|
| Cube | The result of multiplying a number by itself three times, e.g. $2^3 = 8$ |
| Perfect cube | A number that is the cube of some natural number |
| Cube root | The number whose cube gives the given number; symbol $\sqrt[3]{\ }$ |
| Prime factor | A factor that is a prime number |
| Prime factorisation | Expressing a number as a product of its prime factors |
| Square number | A number that is the square of some natural number |
| Even number | A number divisible by 2 |
| Odd number | A number not divisible by 2 |
| Unit place | The right-most digit position of a number |
| Method of estimation | Grouping digits to find a cube root approximately |
| Ramanujan number | 1729 — the smallest number expressible as a sum of two cubes in two ways |