NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots – Here are all the NCERT solutions for Class 8 Maths Chapter 7. This solution contains questions, answers, images, explanations of the complete chapter 7 titled Cubes and Cube Roots of Maths taught in class 8. If you are a student of class 8 who is using NCERT Textbook to study Maths, then you must come across chapter 7 Cubes and Cube Roots. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots in one place.
NCERT Solutions Class 8 Maths Chapter 7 Cubes and Cube Roots
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| Class | 8 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 7 |
| Chapter Name |
Cubes and Cube Roots |
NCERT Solutions Class 8 Maths chapter 7 Cubes and Cube Roots
Class 8, Maths chapter 7, Cubes and Cube Roots solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Find the one’s digit of the cube of each of the following numbers.
(i) 3331
(ii) 8888
(iii) 149
(iv) 1005
(v) 1024
(vi) 77
(vii) 5022
(viii) 53
Ans : Missing
Q.2: Express the following numbers as the sum of odd numbers using the above pattern?
(a) \( 6^{3} \)
(b) \( 8^{3} \)
(c) \( 7^{3} \)
Consider the following pattern.
\( 2^{3}-1^{3}=1+2 \times 1 \times 3 \)
\( 3^{3}-2^{3}=1+3 \times 2 \times 3 \)
\( 4^{3}-3^{3}=1+4 \times 3 \times 3 \)
Using the above pattern, find the value of the following.
(i) \( 7^{3}-6^{3} \)
(ii) \( 12^{3}-11^{3} \)
(iii) \( 20^{3}-19^{3} \)
(iv) \( 51^{3}-50^{3} \)
Ans : Missing
Q.3: Which of the following are perfect cubes?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Ans : Missing
Q.4: Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Ans : (i) The prime factorisation of 216 is as follows:\( 216=2 \times 2 \times 2 \times 3 \times 3 \times 3=2^{3} \times 3^{3} \) Here, as each prime factor is appearing as times as a perfect multiple of 3, therefore, 216 is a perfect cube. (ii) The prime factorisation of 128 is as follows:
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 Here, each prime factor is not appearing as many times as a perfect multiple of 3. One is 2 remaining after grouping the triplets of 2, 128 is not a perfect cube. (iii) The prime factorisation of 1000 is as follows:
1000 = 2 x 2 x 2 x 5 x 5 x 5 Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 1000 is a perfect cube. (iv) The prime factorisation of 100 is as follows:
100 = 2 x 2 x 5 x 5 Here, each prime factor is not appearing as many times as a perfect multiple of 3. Two 2s and two 5s are remaining after grcmping the triplets. Therefore, 100 is not a perfect cube. (v) The prime factorisation of 46656 is as follows:
46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3 Here, as each prime factor is appearing as many times as a perfect multiple of 3, therefore, 46656 is a perfect cube.
Q.5: 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
Ans : (i) 243 = 3 x 3 x 3 x 3 x 3 Here, two 3s are left which are not in a triplet. To make 243 a cube, one more 3 is required. In that case, 243 x 3 = 3 x 3 x 3 x 3 x 3 x 3 = 729 is a perfect cube. Hence, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3. (ii) 256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 Here, two 2s are left which are not in a triplet. To make 256 a cube, one more 2 is required. Then, we obtain 256 x 2 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 512 is a perfect cube. Hence, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2. (iii) 72 = 2 x 2 x 2 x 3 x 3 Here, two 3s are left which are not in a triplet. To make 72 a cube, one more 3 is required. Then, we obtain, 72 x 3 = 2 x 2 x 2 x 3 x 3 x 3 = 216 is a perfect cube. Hence, the smallest natural number by which 72 should be multiplied to make it a perfect cube is 3. (iv) 675 = 3 x 3 x 3 x 5 x 5 Here, two 5s are left which are not in triplet. To make 675 a cube, one more 5 is required. Then, we obtain 675 x 5 = 3 x 3 x 3 x 5 x 5 x 5 = 3756 is a perfect cube. Hence , the smallest natural number by which 675 should be multiplied to make it a perfect cube is 5. (v) 100 = 2 x 2 x 5 x 5 Here, two 2s and two 5s are left which are not in a triplet. To make 100 a cube, we require one more 2 and one more 5. Then, we obtain 100 x 2 x 5 = 2 x 2 x 2 x 5 x 5 x 5 = 1000 is a perfect cube. Hence, the smallest natural number by which 100 should be multiplied to make it a perfect cube is 2 x 5 = 10
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