RS Aggarwal Solutions for Class 8 Maths Chapter 4- Cubes and Cube Roots, are provided here. Our expert faculty team has prepared solutions to help you with your exam preparation to attain good marks in Maths. RS Aggarwal Solutions for Class 8 Maths comes in very handy at this point. If you wish to secure an excellent score, solving RS Aggarwal Class 8 Solutions is an utmost necessity. Solutions that are provided here will help you in getting acquainted with a wide variation of questions and thus, develop problem-solving skills. Practising the textbook questions will help you in analyzing your level of preparation and knowledge of the concept.

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## Exercise 4A

**1.Evaluate :**

**i. (8) ^{3}**

Solution: To evaluate the cube of (8)^{3}

We need to multiply the given number three times i.e. 8 Ã— 8 Ã— 8 = 512

âˆ´ the cube of 8 is 512

**ii. (15) ^{3}**

Solution: To evaluate the cube of (15)^{3}

We need to multiply the given number three times i.e. 15 Ã— 15 Ã— 15 = 3375

âˆ´ the cube of 15 is 3375

**iii. (21) ^{3}**

Solution: To evaluate the cube of (21)^{3}

We need to multiply the given number three times i.e. 21 Ã— 21 Ã— 21 = 9261

âˆ´ the cube of 21 is 9261

**iv. (60) ^{3}**

Solution: To evaluate the cube of (60)^{3}

We need to multiply the given number three times i.e. 60 Ã— 60 Ã— 60 = 216000

âˆ´ the cube of 60 is 216000

**2. Evaluate:**

**i. (1.2) ^{3}**

Solution: To evaluate the cube of (1.2)^{3}

We need to multiply the given number three times i.e. 1.2 Ã— 1.2 Ã— 1.2 = 1.728

We now have to convert the given number into fraction we get,

(1728/1000) = 216/125

âˆ´ the cube of 1.2 is 1.728

**ii. (3.5) ^{3}**

Solution: To evaluate the cube of (3.5)^{3}

We need to multiply the given number three times i.e. 3.5 Ã— 3.5 Ã— 3.5 = 42.875

We now have to convert the given number into fraction we get,

(42875/1000) = 343/8

âˆ´ the cube of 3.5 is 42.875

**iii. (0.8) ^{3}**

Solution: To evaluate the cube of (0.8)^{3}

We need to multiply the given number three times i.e. 0.8 Ã— 0.8 Ã— 0.8 = 0.512

We now have to convert the given number into fraction we get,

(512/1000) = 64/125

âˆ´ the cube of 1.2 is 0.512

**iv. (0.05) ^{3}**

Solution: To evaluate the cube of (0.05)^{3}

We need to multiply the given number three times i.e. 0.05 Ã— 0.05 Ã— 0.05 = 0.000125

We now have to convert the given number into fraction we get,

(125/1000000) = 1/8000

âˆ´ the cube of 1.2 is 0.000125

**Q3. Evaluate:**

**i. (4/7) ^{3}**

Solution: To evaluate the cube of (4/7)^{3}

We need to multiply the given number three times i.e. (4/7) Ã— (4/7) Ã— (4/7) = (64/343)

âˆ´ the cube of (4/7) is (64/343)

**ii. (10/11) ^{3}**

Solution: To evaluate the cube of (10/11)^{3}

We need to multiply the given number three times i.e. (10/11) Ã— (10/11) Ã— (10/11) = (1000/1331)

âˆ´ the cube of (10/11) is (1000/1331)

**iii. (1/15) ^{3}**

Solution: To evaluate the cube of (1/15)^{3}

We need to multiply the given number three times i.e. (1/15) Ã— (1/15) Ã— (1/15) = (1/3375)

âˆ´ the cube of (1/15) is (1000/3375)

**iv. (1 3/10) ^{3}**

Solution: To evaluate the cube of (1 3/10)^{3}

Firstly we need to convert into proper fraction i.e. (13/10)^{3}

We need to multiply the given number three times i.e. (13/10) Ã— (13/10) Ã— (13/10) = (2197/1000)

âˆ´ the cube of (1 3/10) is (2197/1000)

**Q4. Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.**

**i. 125**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

125 = 5 Ã— 5 Ã— 5

Its cube can be expressed as 5^{3}

âˆ´ 125 is a perfect cube.

**ii. 243**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

243 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

Since the given number has more than three factors

âˆ´ 243 is not a perfect cube.

**iii. 343**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

343 = 7 Ã— 7 Ã— 7

Its cube can be expressed as 7^{3}

âˆ´ 343 is a perfect cube.

**iv. 256**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

256 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2Ã— 2 Ã— 2 Ã— 2

Since the given number has more than three factors

âˆ´ 256 is not a perfect cube.

**v. 8000**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

8000 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2Ã— 2 Ã— 5 Ã— 5 Ã— 5

As we can group the above factors into three groups of cubes i.e. 2^{3}, 2^{3}, 5^{3}

We express it as, 2^{3 }Ã— 2^{3} Ã— 5^{3 }= 20^{3}

âˆ´ 8000 is a perfect cube.

**vi. 9261**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

8000 = 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7Ã— 7

As we can group the above factors into three groups of cubes i.e. 3^{3}, 7^{3}

We express it as, 3^{3 }Ã— 7^{3} = 21^{3}

âˆ´ 9261 is a perfect cube.

**vii. 5324**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

5324 = 2 Ã— 2 Ã— 11 Ã— 11 Ã— 11

Since the given number does not have cubical value

âˆ´ 5324 is not a perfect cube.

**viii. 3375**

Solution: A perfect cube can be expressed as a product of three numbers of equal factors

Now by resolving the given number into prime factors we get,

3375 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 5

As we can group the above factors into three groups of cubes i.e. 3^{3}, 5^{3}

We express it as, 3^{3 }Ã— 5^{3} = 15^{3}

âˆ´ 3375 is a perfect cube.

**Q5.Which of the following are the cubes of even numbers?**

**i. 216**

**ii. 729**

**iii. 512**

**iv. 3375**

**v. 1000**

Solution: we know by the rule of even numbers the cubes of even numbers are always even.

So we look for the even numbers in the given set of numbers

216, 512, 1000 are the even numbers

Now we find the prime factors for 216, 512, 1000

216 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 = 2^{3} Ã— 3^{3 }= 6^{3}

512 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 2^{3} Ã— 2^{3} Ã— 2^{3 } = 8^{3}

1000 = 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5 = 2^{3} Ã— 5^{3} = 10^{3}

âˆ´ 216, 512, 1000 are cubes of even numbers.

## Exercise 4B

Find the value of each of the following using the short-cut method:

**i. (25) ^{3}**

Solution: By using the formula (a + b)^{ 3}= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

a^{2} |
a^{2} |
b^{2} |
b^{2} |

Ã—a | Ã—3b | Ã—3a | Ã—b |

a^{3} |
3 a^{2}b |
3ab^{2} |
b^{3} |

Similarly, let us consider a = 2 and b = 5

4 | 4 | 25 | 25 |

Ã—2 | Ã—15 | Ã—6 | Ã—5 |

8 | 60 | 150 | 125 |

+7 | +16 | +12 | |

15 |
76 |
162 |

By considering the last unit digit position (highlighted) numbers we get,

(25)^{3} = 15625

**ii. (47) ^{3}**

Solution: By using the formula (a + b)^{ 3}= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

a^{2} |
a^{2} |
b^{2} |
b^{2} |

Ã—a | Ã—3b | Ã—3a | Ã—b |

a^{3} |
3 a^{2}b |
3ab^{2} |
b^{3} |

Similarly, let us consider a = 4 and b = 7

16 | 16 | 49 | 49 |

Ã—4 | Ã—21 | Ã—12 | Ã—7 |

64 | 336 | 588 | 343 |

+39 | +62 | +34 | |

103 |
398 |
622 |

By considering the last unit digit position (highlighted) numbers we get,

(47)^{3} = 103823

**iii. (68) ^{3}**

Solution: By using the formula (a + b)^{ 3}= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

a^{2} |
a^{2} |
b^{2} |
b^{2} |

Ã—a | Ã—3b | Ã—3a | Ã—b |

a^{3} |
3 a^{2}b |
3ab^{2} |
b^{3} |

Similarly, let us consider a = 6 and b = 8

36 | 36 | 64 | 64 |

Ã—6 | Ã—24 | Ã—18 | Ã—8 |

216 | 864 | 1152 | 512 |

+98 | +120 | +51 | |

314 |
984 |
1203 |

By considering the last unit digit position (highlighted) numbers we get,

(68)^{3} = 314432

**iv. (84) ^{3}**

Solution: By using the formula (a + b)^{ 3}= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

a^{2} |
a^{2} |
b^{2} |
b^{2} |

Ã—a | Ã—3b | Ã—3a | Ã—b |

a^{3} |
3 a^{2}b |
3ab^{2} |
b^{3} |

Similarly, let us consider a = 8 and b = 4

64 | 64 | 16 | 16 |

Ã—8 | Ã—12 | Ã—24 | Ã—4 |

512 | 768 | 384 | 64 |

+80 | +39 | +6 | |

592 |
807 |
390 |

By considering the last unit digit position (highlighted) numbers we get,

(84)^{3} = 592704

## Exercise 4C

**Evaluate**

**1. âˆ›64**

Solution: The prime factors of 64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

Now by grouping into three we get, (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2)

âˆ´ âˆ› ((2)^{3} Ã— (2)^{3}) = (2 Ã— 2) = 4

**2. âˆ›343**

Solution: The prime factors of 343 = 7 Ã— 7 Ã— 7

Now by grouping into three we get, (7 Ã— 7 Ã— 7)

âˆ´ âˆ› (7)^{3} = 7

**3. âˆ›729**

Solution: The prime factors of 729 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

Now by grouping into three we get, (3 Ã— 3 Ã— 3) Ã— (3 Ã— 3 Ã— 3)

âˆ´ âˆ› ((3)^{3} Ã— (3)^{3}) = (3 Ã— 3) = 9

**4. âˆ›1728**

Solution: The prime factors of 1728 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

Now by grouping into three we get, (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2) Ã— (3 Ã— 3 Ã— 3)

âˆ´ âˆ› ((2)^{3} Ã— (2)^{3}Ã— (3)^{3}) = (2 Ã— 2 Ã— 3) = 12

**5. âˆ›9261**

Solution: The prime factors of 9261 = 3 Ã— 3 Ã— 3 Ã— 7 Ã— 7 Ã— 7

Now by grouping into three we get, (3 Ã— 3 Ã— 3) Ã— (7 Ã— 7 Ã— 7)

âˆ´ âˆ› ((3)^{3} Ã— (7)^{3}) = (3 Ã— 7) = 21

**6. âˆ›4096**

Solution: The prime factors of 4096 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

Now by grouping into three we get, (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2)

âˆ´ âˆ› ((2)^{3} Ã— (2)^{3}Ã— (2)^{3} Ã— (2)^{3}) = (2 Ã— 2 Ã— 2 Ã— 2) = 16

**7. âˆ›8000**

Solution: The prime factors of 8000 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 5

Now by grouping into three we get, (2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2) Ã— (5 Ã— 5 Ã— 5)

âˆ´ âˆ› ((2)^{3} Ã— (2)^{3}Ã— (5)^{3}) = (2 Ã— 2 Ã— 5) = 20

**8. âˆ›3375**

Solution: The prime factors of 3375 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 5

Now by grouping into three we get, (3 Ã— 3 Ã— 3) Ã— (5 Ã— 5 Ã— 5)

âˆ´ âˆ› ((3)^{3} Ã— (5)^{3}) = (3 Ã— 5) = 15

**9. âˆ›-216**

Solution: The prime factors of 216 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

Now by grouping into three we get, (2 Ã— 2 Ã— 2) Ã— (3 Ã— 3 Ã— 3)

âˆ´ âˆ› – ((2)^{3} Ã— (3)^{3}) = – (2 Ã— 3) = -6

## Exercise 4D

**Tick the correct answer in each of the following:**

**1. Which of the following numbers is a perfect cube?**

**a) 141**

**b) 294**

**c) 216**

**d) 496**

Solution: Firstly we have to find the factors for the above numbers

âˆ›141 = âˆ› (3 Ã— 47)

âˆ›294 = âˆ› (2 Ã— 3 Ã— 7 Ã— 7)

âˆ›216 = âˆ› (6 Ã— 6 Ã— 6)

âˆ›496 = âˆ› (2 Ã— 2 Ã— 2 Ã— 2 Ã— 31)

From the above results, 216 has the perfect cube factors.

âˆ´ 216 is the perfect cube.

**2. Which of the following numbers is a perfect cube?**

**a) 1152**

**b) 1331**

**c) 2016**

**d) 739**

Solution: Firstly we have to find the factors for the above numbers

âˆ›1152 = âˆ› (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3) = 4 âˆ› (2 Ã— 3 Ã— 3)

âˆ›1331 = âˆ› (11 Ã— 11 Ã— 11)

âˆ›2016 = âˆ› (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7) = 2 âˆ› (2 Ã— 2 Ã— 3 Ã— 3 Ã— 7)

âˆ›739 = no prime factors can be found.

From the above results, 1331 has the perfect cube factors.

âˆ´ 1331 is the perfect cube.

**3. âˆ›512 =?**

**a) 6**

**b) 7**

**c) 8**

**d) 9**

Solution: Firstly we have to find the prime factors of 512

âˆ›512 = âˆ› (8 Ã— 8 Ã— 8) = 8

âˆ´ 8 is the correct answer.

**4. âˆ› (125 Ã— 64) =?**

**a) 100**

**b) 40**

**c) 20**

**d) 30**

Solution: Firstly we have to find the prime factors of 125

125 = 5 Ã— 5 Ã— 5

And the prime factors of 64

64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

âˆ› (125 Ã— 64) = âˆ› (5 Ã— 5 Ã— 5 Ã— 4 Ã— 4 Ã— 4) = âˆ› (5 Ã— 5 Ã— 5) Ã— âˆ› (4 Ã— 4 Ã— 4) = 5 Ã— 4 = 20

âˆ´ 20 is the correct answer.

**5. âˆ› (64/343) =?**

**a) 4/9**

**b) 4/7**

**c) 8/7**

**d) 8/21**

Solution: Firstly we have to find the prime factors of 64

64 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

And the prime factors of 343

343 = 7 Ã— 7 Ã— 7

âˆ› (64/343) = âˆ› (4 Ã— 4 Ã— 4/7 Ã— 7 Ã— 7)

âˆ› (64/343) = 4/7

## RS Aggarwal Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots

Chapter 4- Cubes and Cube Roots contain 4 exercises and the RS Aggarwal solutions present in this page provide the answers for the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Cube of a number – A natural number is said to be a perfect cube if it is the cube of some natural number.
- Cubes of Negative Integers.
- Cube of a Rational Number.
- Properties of Cubes of Numbers.

### Also, Access RS Aggarwal Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots Exercises

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots

RS Aggarwal Solutions for Class 8 Maths Chapter 4 – Cubes and Cube Roots ensures that the students are thorough and familiar with the concepts. The solutions are designed in such a way that they are easy to understand and solve. Cube roots are used in day to day Mathematics like in powers and exponents or to find the side of a three-dimensional cube when its volume is given. In RS Aggarwal Solutions, many such exercise problems are solved which enhances familiarity with the concepts.