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In our childhood, many of us have tried to stack small cubes to create a bigger cube box. Also, some of you have already come across many perfect shapes-related puzzles like Rubik's Cube. These are all based on the idea of Cubes. In mathematics, when you multiply a number by itself three times, then the result is the cube of that number. For example cube of 2 is (2 × 2 × 2) = 8. Also, the cube root is the number which is used to make a cube, i.e. cube root of 8 is 2. In the 6th chapter of the NCERT Class 8 Maths, you will find Cubes and Cube Roots.
This article on NCERT solutions for class 8 Maths Chapter 6 Cubes and Cube Roots offers clear and step-by-step solutions for the exercise problems in the NCERT Books for class 8 Maths. Students who are in need of Cubes and Cube Roots class 8 solutions will find this article very useful. It covers all the important Class 8 Maths Chapter 6 question answers. These Cubes and Cube Roots class 8 ncert solutions are made by the Subject Matter Experts according to the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. NCERT solutions for class 8 maths and NCERT solutions for other subjects and classes can be downloaded from the NCERT Solutions.
Cube Root Formula: For any number $m$, which can be expressed as the product of any number three times, $m = n × n × n = n^3$
The cube root of $m$ is denoted as: $\sqrt[3]{m}=n$
Methods of Finding Cube Roots:
Prime Factorization Method
Estimation Method
Class 8 maths chapter 6 question answer - Topic 6.2 Cubes
Q(i) Find the one’s digit of the cube of each of the following numbers.
3331
Answer: The detailed solution for the above-mentioned question is as follows,
Since the given number ends with 1, so the one’s digit of the cube of 3331 will be 1.
Q(ii) Find the one’s digit of the cube of each of the following numbers.
8888
Answer: The detailed solution for the above-mentioned question is as follows
Since the given number ends with 8, so the one’s digit of the cube of 8888 will be 2.
Q(iii) Find the one's digit of the cube of each of the following numbers.
149
Answer: The detailed solution for the above-mentioned question is as follows,
Since the given number has 9 at units place, so the one’s digit of the cube of 149 will be 9.
Q(iv) Find the one’s digit of the cube of each of the following numbers.
1005
Answer: The detailed solution for the above-mentioned questions is as follows
Since the given number ends with 5, so one's digit of its cube will also end with 5.
Q(v) Find the one’s digit of the cube of each of the following numbers.
1024
Answer: The solution to the above-mentioned question is as follows,
The given digit ends with 4. So the one’s digit of the cube of 1024 will be 4.
Q(vi) Find the one’s digit of the cube of each of the following numbers.
77
Answer: The detailed solution for the above-mentioned question is as follows,
The given number ends with 7, so its cube will end with 3.
Q(vii) Find the one’s digit of the cube of each of the following numbers.
5022
Answer: The detailed solution for the above-mentioned question is as follows,
Since the given number ends with 2, its cube will end with 8.
Q(viii) Find the one’s digit of the cube of each of the following numbers.
53
Answer: The detailed solution for the above-mentioned question is as follows,
Since the given number has 3 at units place, so, its cube will end with 7.
NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots: Topic 6.2.1
Subtopic Some Interesting Patterns
Q(a) Express the following numbers as the sum of odd numbers using the above pattern?
$6^{3}$
Answer: The detailed solution for the above-mentioned question is as follows,
$6^3 =$ 216 => 31 + 33 + 35 + 37 + 39 + 41
Q(b) Express the following numbers as the sum of odd numbers using the above pattern?
$8^{3}$
Answer: The detailed solution for the above-mentioned question is as follows
$8^3 =$ 512 => 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71
Q(c) Express the following numbers as the sum of odd numbers using the above pattern?
$7^{3}$
Answer: The detailed solution for the above-mentioned question is as follows
73 = 43 + 45 + 47 + 49 + 51 + 53 + 55
Q(i) The detailed solution of the above-written question is as follows,
Using the above pattern
$2^{3}-1^{3}=1+2\times 1\times 3$
$3^{3}-2^{3}=1+3\times2\times 3$
$4^{3}-3^{3}=1+4\times 3\times 3$
Find the value of the following
$7^{3}-6^{3}$
Answer: The value of the following question is:
$7^{3}-6^{3} =$ $7^{3}-6^{3} = 1 + 7\times6\times3 = 1 + 126 = 127$
Q(ii) The detailed solution for all the above-written questions is as follows
Using the above pattern
$2^{3}-1^{3}=1+2\times 1\times 3$
$3^{3}-2^{3}=1+3\times2\times 3$
$4^{3}-3^{3}=1+4\times 3\times 3$
Find the value of the following.
$12^{3}-11^{3}$
Answer: $12^{3}-11^{3} = 1 + 12\times11\times3 = 1 + 396 = 397$
Q(iii). Consider the following pattern.
$2^{3}-1^{3}=1+2\times 1\times 3$
$3^{3}-2^{3}=1+3\times2\times 3$
$4^{3}-3^{3}=1+4\times 3\times 3$
Using the above pattern, find the value of the following.
$20^{3}-19^{3}$
Answer: The detailed solution for the above-written question is mentioned below,
$20^{3}-19^{3} = 1 + 20\times19\times3 = 1 + 1140 = 1141$
Q(iv) Consider the following pattern.
$2^{3}-1^{3}=1+2\times 1\times 3$
$3^{3}-2^{3}=1+3\times2\times 3$
$4^{3}-3^{3}=1+4\times 3\times 3$
Using the above pattern, find the value of the following.
$51^{3}-50^{3}$
Answer: The detailed solution for the above question is mentioned below
$51^{3}-50^{3} = 1 + 51\times50\times3 = 1 + 7650 = 7651$
NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots Topic 6.2.1
Subtopic Cubes and Their Prime Factors
Q. Which of the following are perfect cubes?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Answer: We will find it by prime factorization whether they make a pair of three prime numbers or not.
(1) $400 = 2\times2\times2\times2\times5\times5$ . So not a perfect cube.
(2) $3375 = 3\times3\times3\times5\times5\times5$ . So it is a perfect cube.
(3) $8000 = 2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is a perfect cube.
(4) $15625 = 5\times5\times5\times5\times5\times5$ . So it is a perfect cube.
(5) $9000 = 2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.
(6) $6859 = 19\times19\times19$ . So it is a perfect cube.
(7) $2025 = 3\times3\times3\times3\times5\times5$ . So it is not a perfect cube.
(8) $10648 = 2\times2\times2\times11\times11\times11$ . So it is a perfect cube.
NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots
Topic 6.2.2 Smallest Multiple That is a Perfect Cube
Q. Check which of the following are perfect cubes.
(i) 2700
(ii) 16000
(iii) 64000
(iv) 900
(v) 125000
(vi) 36000
(vii) 21600
(viii) 10,000
(ix) 27000000
(x) 1000.
What pattern do you observe in these perfect cubes?
Answer: The detailed solution for the above-written question is as follows
By prime factorization:
(i) $2700 = 2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.
(ii) $16000 = 2\times2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is not a perfect cube.
(iii) $64000 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times5\times5\times5= 80\times80\times80$ . So it is a perfect cube.
(iv) $900 = 2\times2\times3\times3\times5\times5$ . So it is not a perfect cube.
(v) $125000 = 2\times2\times2\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.
(vi) $36000 = 2\times2\times2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.
(vii) $21600 = 2\times2\times2\times2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.
(viii) $10000 = 2\times2\times2\times2\times5\times5\times5\times5$ . So it is not a perfect cube.
(ix) $27000000 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.
(x) $1000 = 2\times2\times2\times5\times5\times5$ . So it is a perfect cube.
We observe that the numbers above which are perfect cube have the number of zeros in multiple of 3.
NCERT Class 8 Solution Exercise: 6.1
Page number: 76, Total questions: 4
Q.1(i) Which of the following numbers are not perfect cubes?
216
Answer: The detailed solution for the above-written question is as follows
By prime factorization of 216 gives:
$216 = 2\times2\times2\times3\times3\times3$
Since prime numbers are present in pairs of three, so the given number is a perfect cube.
Q.1(ii) Which of the following numbers are not perfect cubes?
128
Answer: We have 128. By prime factorization we get,
$128 = 2\times2\times2\times2\times2\times2\times2$
Since the prime numbers are not in pairs of three, so the given number is not a perfect cube.
Q.1(iii) Which of the following numbers are not perfect cubes?
1000
Answer: The detailed solution for the above written is as follows
By prime factorization of 1000 we get :
$1000 = 2\times2\times2\times5\times5\times5$ .
So the given number is a perfect cube.
Q.1(iv) Which of the following numbers are not perfect cubes?
100
Answer: The detailed solution for the above-written question is as follows
By prime factorization of 100 :
$100 = 2\times2\times5\times5$ .
Since prime numbers are not in pair of three so given number is not a perfect cube.
Q.1(v) Which of the following numbers are not perfect cubes?
46656
Answer: We have 46656, by prime factorisation:
$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$ .
Since prime numbers are in group of three. So the given number is a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100
Answer: This can be found by knowing about the prime factors of the number.
(i) 243 : $3\times3\times3\times3\times3$ .
So it must be multiplied by 3.
(ii) 256 : $2\times2\times2\times2\times2\times2\times2\times2$
So the given number must be multiplied by 2 to make it a perfect cube.
(iii) 72 : $2\times2\times2\times3\times3$
So 72 must be multiplied by 3 to make it a perfect cube.
(iv) 675 : $3\times3\times3\times5\times5$
So it should be multiplied by 5.
(v) 100 : $2\times2\times5\times5$
So it should be multiplied by 10.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704
Answer: By prime factorization of given numbers:
(i) 81: $3\times3\times3\times3$
So given number needs to be divided by 3 to get a perfect cube.
(ii) 128: $2\times2\times2\times2\times2\times2\times2$ .
So the given number needs to be divided by 2 to get a perfect cube.
(iii) 135: $3\times3\times3\times5$
So the given number needs to be divided by 5 to get a perfect cube.
(iv) 192: $2\times2\times2\times2\times2\times2\times3$
So the given number needs to be divided by 3 to get a perfect cube.
(v) 704: $2\times2\times2\times2\times2\times2\times11$
So the given number needs to be divided by 11 to get a perfect cube.
Answer: Volume of cuboid is $5\times2\times5 = 2\times5\times5$ $cm^3$
To make it a cube need to make this a perfect cube number.
So we need $2\times2\times5$ cuboids or 20 cuboids.
for any integer $m,m^{2}< m^{3}.$ Why?
Answer: The detailed solution for the above-written question is as follows.
False.
$m^2 < m^3$
Or, $0 < m^3 - m^2$
Or, $m^3 - m^2>0$
Or $m^2\left ( m-1 \right )>0$
Now, put any number less than 1, we see that this relation doesn't hold.
So for m<1 this condition is not true.
NCERT Class 8 Solution Exercise: 6.2
Page number: 77, Total questions: 2
Q.1(i) Find the cube root of each of the following numbers by the prime factorisation method.
64
Answer: The detailed solution for the above-written question is as follows
Prime factorization of 64 gives :
$64 = 2\times2\times2\times2\times2\times2$
So its cube root is $2\times2$ = 4
Q.1(ii) Find the cube root of each of the following numbers by prime factorisation method.
512
Answer: By prime factorisation of 512:
$512 = 2\times2\times2\times2\times2\times2\times2\times2\times2$
So its cube root is $2\times2\times2 = 8$
Q.1(iii) Find the cube root of each of the following numbers by the prime factorisation method.
10648
Answer: The detailed solution for the above-written question is as follows
Prime factorization of 10648 gives :
$10648 = 2\times2\times2\times11\times11\times11$
So, its cube root is 22.
Q.1(iv) Find the cube root of each of the following numbers by the prime factorisation method.
27000
Answer: The detailed solution for the above-written question is as follows
By the prime factorization method, we get :
$27000 = 2\times2\times2\times3\times3\times3\times5\times5\times5$
So its cube root is 30.
Q.1(v) Find the cube root of each of the following numbers by the prime factorisation method.
15625
Answer: The detailed solution for the above-written question is as follows
By prime factorization:
$15625 = 5\times5\times5\times5\times5\times5$
So its cube root is 25.
Q.1(vi) Find the cube root of each of the following numbers by prime factorisation method.
13824
Answer: The detailed solution for the above-written question is as follows
By prime factorization:
$13824 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$
So its cube root is 24.
Q.1(vii) Find the cube root of each of the following numbers by the prime factorisation method.
110592
Answer: The detailed solution for the above-written question is as follows
By prime factorization:
$110592 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$
So its cube root is $2\times2\times2\times2\times3$ = 48.
Q.1(viii) Find the cube root of each of the following numbers by the prime factorisation method.
46656
Answer: The detailed solution for the above-written question is as follows
By prime factorization, we get :
$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$
So its cube root is $2\times2\times3\times3$ = 36.
Q.1(ix) Find the cube root of each of the following numbers by the prime factorisation method.
175616
Answer: The detailed solution for the above-written question is as follows
By prime factorization, we get :
$175616 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7$
So its cube root is $2\times2\times2\times7$ = 56.
Q.1(x) Find the cube root of each of the following numbers by prime factorisation method.
91125
Answer: The detailed solution for the above-written question is as follows
By prime factorization, we get :
$91125 = 3\times3\times3\times3\times3\times3\times5\times5\times5$
So its cube root is $3\times3\times5$ = 45.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If the square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8
(v) The cube of a two-digit number may be a three-digit number.
(vi) The cube of a two-digit number may have seven or more digits.
(vii) The cube of a single-digit number may be a single-digit number.
Answer: (i) False. The cube of an odd number can never be even.
(ii) True. A perfect cube number ends with zeros multiples of three.
(iii) False. We can say only about the unit's place.
(iv) False. Cube of numbers which end with 2 end with 8.
(v) False. It can never be.
(vi) False. It can never be. It can be proved by taking examples.
(vii) True. e.g. 1,2
Students can use the links below to prepare efficiently in other subjects as well as Mathematics.
The following links will give students access to the latest CBSE syllabus and some reference books.
Keep Working Hard and Happy Learning!
Finding cubes and cubes roots for numbers containing up to 3 digits, estimating square roots and cube roots are two important topics of this chapter.
NCERT solutions not only helpful for the students if they stuck while solving NCERT problems but also they will get conceptual clarity as these solutions are provided in a very detailed manner.
No, CBSE doesn't provide NCERT solutions for any class and subject.
Here you will get the detailed NCERT solutions for class 8 by clicking on the link.
Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.
There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.
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