NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Root Exercise 6.1

Exercise 6.1 of Class 8 Maths Chapter 6 helps you understand the basic concept of cube numbers. This exercise will teach you how to find cube of any number, the meaning of cube, how to observe the properties of cubes and recognize the patterns of cube numbers. These aspects of maths are not only helpful in solving maths problems, they also support other subjects like Physics and Algebra.

This whole chapter and exercise are designed based on the latest NCERT syllabus for Class 8, as mentioned in the official textbook by NCERT. Having a proper understanding of cube numbers will help in scoring well during school exams and other competitive exams like the Olympiads

The NCERT Solutions helps to understand how to solve each question in Exercise 6.1 properly. This will improve your accuracy, speed and clarity of concepts, and make you prepared and confident in your examination.

1.0Download NCERT Solutions Class 8 Maths Chapter 6 Cubes and Cube Root Exercise 6.1: Free PDF

Master exercise 6.1 with the free PDF of NCERT Solutions for Class 8 Maths Chapter 6 available here. The Solutions are provided in a step by step manner for clear understanding.

Download the PDF from the link below:

2.0Key Concepts in Exercise 6.1 of Class 8 Maths Chapter 6

This exercise helps students understand cube numbers and how to identify or calculate them. The key concepts included in this exercise are:

• Definition and identification of cube numbers
• Cubes of small natural numbers
• Recognizing patterns in cube numbers
• Properties of cube numbers
• Ending digits of cubes
• Cubes of even and odd numbers

3.0NCERT Class 8 Maths Chapter 6: Other Exercises

4.0NCERT Class 8 Maths Chapter 6 Exercise 6.1: Detailed Solutions

• Which of the following numbers are not perfect cube? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656 Sol. (i) Resolving 216 into prime factors, we find that

$216=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}$ Clearly, the prime factors of 216 can be grouped into triples of equal factors and no factor is left over. $\therefore 216$ is a perfect cube (ii) Resolving 128 into prime factors, we find that

$128=\underline{2\times 2\times 2\times 2\times 2\times 2\times 2}$ Now, if we try to group together triples of equal factors, we are left with a single factor, 2. $\therefore 128$ is not a perfect cube. (iii) Resolving 1000 into prime factors, we find that

$1000=\underline{2\times 2\times 2\times 5\times 5\times 5}$ Clearly, the prime factors of 1000 can be grouped into triples of equal factors and no factor is left over. $\therefore 1000$ is a perfect cube. (iv) Resolving 100 into prime factors, we find that

$100=2\times 2\times 5\times 5$. Now, if we try to group together triples of equal factors, we are left with $2\times 2\times 5\times 5$ $\therefore 100$ is not a perfect cube. (v) Resolving 46656 into prime factors, we find that

$46656=\underline{2\times 2\times 2\times 2\times 2\times 2\times \underline{3\times 3\times 3}\times }$ $\underline{3\times 3\times 3}$ 46656 can be grouped into triples of equal factors and no factor is left over. $\therefore 46656$ is a perfect cube.

• Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 (ii)256 (iii) 72 (iv) 675 (v) 100

Sol. (i) Writing 243 as a product of prime factors, we have

$243=\underline{3\times 3\times 3}\times 3\times 3$ Clearly, to make it a perfect cube, it must be multiplied by 3 . (ii) Writing 256 as a product of prime factors, we have

$256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2\times 2\times 2}$ Clearly, to make it a perfect cube, it must be multiplied by 2 . (iii) Writing 72 as a product of prime factors, we have

$72=\underline{2\times 2\times 2}\times 3\times 3$ Clearly, to make it a perfect cube, it must be multiplied by 3 . (iv) Writing 675 as a product of prime factors, we have

$675=\underline{3\times 3\times 3}\times 5\times 5$ Clearly, to make it a perfect cube, it must be multiplied by 5 . (v) Writing 100 as a product of prime factors, we have

$100=2\times 2\times 5\times 5$ Clearly, to make it a perfect cube, it must be multiplied by $2\times 5=10$.

• Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704 Sol. (i) Writing 81 as a product of prime factors, we have

$81=\underline{3\times 3\times 3\times 3}$

Clearly, to make it a perfect cube, it must be divided by 3 . (ii) Writing 128 as a product of prime factors, we have

$128=\underline{2\times 2\times 2\times 2\times 2\times 2\times 2}$ Clearly to make it a perfect cube, it must be divided by 2 . (iii) Writing 135 as a product of prime factors, we have

$135=\underline{3\times 3\times 3}\times 5$ Clearly, to make it a perfect cube, it must be divided by 5 . (iv) Writing 192 as a product of prime factors, we have

$192=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 3$ Clearly, to make it a perfect cube, it must be divided by 3 . (v) Writing 704 as a product of prime factors, we have

$704=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 11$ Clearly, to make it a perfect cube, it must be divided by 11 .

• Parikshit makes a cuboid of plasticine of side $5\mathrm{cm},2\mathrm{cm},5\mathrm{cm}$. How many such cuboids will he need to form a cube? Sol. Volume of the cuboid $=5\times 2\times 5=2\times 5\times 5$ To make it a cube, we require $2\times 2\times 5$ Hence, volume of cube $=(2\times 5\times 5\times 2\times 2\times 5){\mathrm{cm}}^3$ or $(2\times 2\times 2\times 5\times 5\times 5){\mathrm{cm}}^3$. Number of cuboids $=\frac{\text{ Volume of cube }}{\text{ Volume of cuboid }}$ $=\frac{2\times 2\times 2\times 5\times 5\times 5}{2\times 5\times 5}=2\times 2\times 5$ $=20$ cuboids

5.0Key Features and Benefits of Class 8 Maths Chapter 6 Exercise 6.1

• This exercise covers the essential principle of cubes and introduces it with clear examples.
• The Questions strictly follow the revised NCERT Class 8 Maths syllabus.
• Understanding of this exercise enables students to see cube number patterns via step wise explanation.
• These solutions build number sense and pattern recognition which can help the Math Olympiad preparation.
• Regular practice increases confidence in solving cubes related questions in CBSE board examinations.
• Helps develop a sound in-running foundation of algebra and advanced number theory for higher classes.