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Ncert solutions for class 8 maths chapter 7 Cube and Cube Root. Here We learn what is in ncert maths class 8 EXERCISE 7 and how to solve questions with easiest method.  In this chapter we solve the question of NCERT class 8th maths EXERCISE 7. NCERT class 8 maths solutions Cube and Cube Root are part of NCERT Solutions for class 8 maths chapter 7 solution PDF. Ncert solutions for class 8 EXERCISE 7 with formula and solution. class 8th maths

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NCERT Solutions for Class 8 Maths EXERCISE 7

Cube and Cube Root

NCERT Solutions for Class 8 Maths chapter 7
class 8th maths
Ex 7.1

EXERCISE 7.1

1. Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Solution :-
(i) 216
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|216}}\\\underline{{\left. 2 \right|108}}\\\underline{{\left. 2 \right|54}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 216 = 2 × 2 × 2 × 3 × 3 × 3 
There are each prime factor appears three times, so number 216 is a perfect cube. 

(ii) 128
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|128}}\\\underline{{\left. 2 \right|64}}\\\underline{{\left. 2 \right|32}}\\\underline{{\left. 2 \right|16}}\\\underline{{\left. 2 \right|8}}\\\underline{{\left. 2 \right|4}}\\\underline{{\left. 2 \right|2}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 
In the above factorisation 3 remains after grouping the 2’s in triplets. Therefore, 128 is not a perfect cube.

(iii) 1000
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|1000}}\\\underline{{\left. 2 \right|500}}\\\underline{{\left. 2 \right|250}}\\\underline{{\left. 5 \right|125}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 1000 = 2 × 2 × 2 × 5 × 5 × 5 
There are each prime factor appears three times, so number 1000 is a perfect cube. 

(iv) 100
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|100}}\\\underline{{\left. 2 \right|50}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 100 = 2 × 2 × 5 × 5 
There are each prime factor do not appears three times, so number 100 is a perfect cube. 

(v) 46656
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|46656}}\\\underline{{\left. 2 \right|23328}}\\\underline{{\left. 2 \right|11664}}\\\underline{{\left. 2 \right|5832}}\\\underline{{\left. 2 \right|2916}}\\\underline{{\left. 2 \right|1458}}\\\underline{{\left. 3 \right|729}}\\\underline{{\left. 3 \right|243}}\\\underline{{\left. 3 \right|81}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
There are each prime factor appears three times, so number 46656 is a perfect cube. 

2. 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

Solution :-
(i) 243
\(\displaystyle \begin{array}{l}\underline{{\left. 3 \right|243}}\\\underline{{\left. 3 \right|81}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 243 = 3 × 3 × 3 × 3 × 3
In this factorisation, we find that there is no triplet of 3.
So, To make it a perfect cube we multiply it by 3.
= 3 × 3 × 3 × 3 × 3 × 3 
= 3 × 3
= 9
Multiplying the number 243 by 3 will get the perfect cube number 729. whose cube root is 9.

(ii) 256
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|256}}\\\underline{{\left. 2 \right|128}}\\\underline{{\left. 2 \right|64}}\\\underline{{\left. 2 \right|32}}\\\underline{{\left. 2 \right|16}}\\\underline{{\left. 2 \right|8}}\\\underline{{\left. 2 \right|4}}\\\underline{{\left. 2 \right|2}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
In this factorisation, we find that there is no triplet of 2.
So, To make it a perfect cube we multiply it by 2.
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2 × 2 × 2
= 8
Multiplying the number 256 by 2 will get the perfect cube number 512. whose cube root is 8.

(iii) 72
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|72}}\\\underline{{\left. 2 \right|36}}\\\underline{{\left. 2 \right|18}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 72 = 2 × 2 × 2 × 3 × 3
In this factorisation, we find that there is no triplet of 3.
So, To make it a perfect cube we multiply it by 3.
= 2 × 2 × 2 × 3 × 3 × 3
= 2 × 3
= 6
Multiplying the number 72 by 3 will get the perfect cube number 216. whose cube root is 6.

(iv) 675
\(\displaystyle \begin{array}{l}\underline{{\left. 5 \right|675}}\\\underline{{\left. 5 \right|135}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 675 = 5 × 5 × 3 × 3 × 3
In this factorisation, we find that there is no triplet of 5.
So, To make it a perfect cube we multiply it by 5.
= 5 × 5 × 5 × 3 × 3 × 3
= 5 × 3
= 15
Multiplying the number 6752 by 5 will get the perfect cube number 3375. whose cube root is 15.

(v) 100
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|100}}\\\underline{{\left. 2 \right|50}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 100 = 2 × 2 × 5 × 5
In this factorisation, we find that there is no triplet of 2 and 5.
So, To make it a perfect cube we multiply it by 2 and 5.
= 2 × 2 × 2 × 5 × 5 × 5
= 2 × 5
= 10
Multiplying the number 100 by 10 will get the perfect cube number 1000. whose cube root is 10.

3. 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

Solution :
(i) 81
\(\displaystyle \begin{array}{l}\underline{{\left. 3 \right|81}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 81 = 3 × 3 × 3 × 3
In this factorisation, we find that there is no triplet of 3.
So, To make it a perfect cube we divide it by 3.
= 3 × 3 × 3
= 3
Divided the number 81 by 3 will get the perfect cube number 27. whose cube root is 3.

(ii) 128 
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|128}}\\\underline{{\left. 2 \right|64}}\\\underline{{\left. 2 \right|32}}\\\underline{{\left. 2 \right|16}}\\\underline{{\left. 2 \right|8}}\\\underline{{\left. 2 \right|4}}\\\underline{{\left. 2 \right|2}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
In this factorisation, we find that there is no triplet of 2.
So, To make it a perfect cube we divide it by 2.
= 2 × 2 × 2 × 2 × 2 × 2
= 2 × 2
= 4
Divided the number 128 by 2 will get the perfect cube number 64. whose cube root is 4.

(iii) 135
\(\displaystyle \begin{array}{l}\underline{{\left. 3 \right|135}}\\\underline{{\left. 3 \right|45}}\\\underline{{\left. 3 \right|15}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 135 = 3 × 3 × 3 × 5
In this factorisation, we find that there is no triplet of 5.
So, To make it a perfect cube we divide it by 5.
= 3 × 3 × 3
= 3
Divided the number 135 by 5 will get the perfect cube number 27. whose cube root is 3.

(iv) 192
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|192}}\\\underline{{\left. 2 \right|96}}\\\underline{{\left. 2 \right|48}}\\\underline{{\left. 2 \right|24}}\\\underline{{\left. 2 \right|12}}\\\underline{{\left. 2 \right|6}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
In this factorisation, we find that there is no triplet of 3.
So, To make it a perfect cube we divide it by 3.
= 2 × 2 × 2 × 2 × 2 × 2
= 2 × 2
= 4
Divided the number 192 by 3 will get the perfect cube number 64. whose cube root is 4.

(v) 704
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|704}}\\\underline{{\left. 2 \right|352}}\\\underline{{\left. 2 \right|176}}\\\underline{{\left. 2 \right|88}}\\\underline{{\left. 2 \right|44}}\\\underline{{\left. 2 \right|22}}\\\underline{{\left. {11} \right|11}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11
In this factorisation, we find that there is no triplet of 11.
So, To make it a perfect cube we divide it by 11.
= 2 × 2 × 2 × 2 × 2 × 2
= 2 × 2
= 4
Divided the number 704 by 11 will get the perfect cube number 64. whose cube root is 4.

4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution : 
Volume of Cuboid = Length × Breadth × Height
= 5 × 2 × 5 cm³
= 2 × 5 × 5 cm³
Number of cuboids needed to make a cube
= 2 × 2 × 5 = 20 cuboids

NCERT Solutions for Class 8 Maths chapter 7
class 8th maths
Ex 7.2

EXERCISE 7.2

1. Find the cube root of each of the following numbers by prime factorisation method.

(i) 64	(ii) 512	(iii) 10648	(iv) 27000
(v) 15625	(vi) 13824	(vii) 110592	(viii) 46656
(ix) 175616	(x) 91125

Solution :
(i) 64
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|64}}\\\underline{{\left. 2 \right|32}}\\\underline{{\left. 2 \right|16}}\\\underline{{\left. 2 \right|8}}\\\underline{{\left. 2 \right|4}}\\\underline{{\left. 2 \right|2}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 64 = 2 × 2 × 2 × 2 × 2 × 2 
So, cube root of 64 = 2 × 2
= 4

(ii) 512
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|512}}\\\underline{{\left. 2 \right|256}}\\\underline{{\left. 2 \right|128}}\\\underline{{\left. 2 \right|64}}\\\underline{{\left. 2 \right|32}}\\\underline{{\left. 2 \right|16}}\\\underline{{\left. 2 \right|8}}\\\underline{{\left. 2 \right|4}}\\\underline{{\left. 2 \right|2}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
So, cube root of 512 = 2 × 2 × 2
= 8

(iii) 10648
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|10648}}\\\underline{{\left. 2 \right|5324}}\\\underline{{\left. 2 \right|2662}}\\\underline{{\left. {11} \right|1331}}\\\underline{{\left. {11} \right|121}}\\\underline{{\left. {11} \right|11}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 10648 = 2 × 2 × 2 × 11 × 11 × 11
So, cube root of 10648 = 2 × 11
= 22

(iv) 27000
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|27000}}\\\underline{{\left. 2 \right|13500}}\\\underline{{\left. 2 \right|6750}}\\\underline{{\left. 3 \right|3375}}\\\underline{{\left. 3 \right|1125}}\\\underline{{\left. 3 \right|375}}\\\underline{{\left. 5 \right|125}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
So, cube root of 27000 = 2 × 3 × 5
= 30

(v) 15625
\(\displaystyle \begin{array}{l}\underline{{\left. 5 \right|15625}}\\\underline{{\left. 5 \right|3125}}\\\underline{{\left. 5 \right|625}}\\\underline{{\left. 5 \right|125}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 15625 = 5 × 5 × 5 × 5 × 5 × 5
So, cube root of 15625 = 5 × 5
= 25

(vi) 13824
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|13824}}\\\underline{{\left. 2 \right|6912}}\\\underline{{\left. 2 \right|3456}}\\\underline{{\left. 2 \right|1728}}\\\underline{{\left. 2 \right|864}}\\\underline{{\left. 2 \right|432}}\\\underline{{\left. 2 \right|216}}\\\underline{{\left. 2 \right|108}}\\\underline{{\left. 2 \right|54}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
So, cube root of 13824 = 2 × 2 × 2 × 3
= 24

(vii) 110592
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|110592}}\\\underline{{\left. 2 \right|55296}}\\\underline{{\left. 2 \right|27648}}\\\underline{{\left. 2 \right|13824}}\\\underline{{\left. 2 \right|6912}}\\\underline{{\left. 2 \right|3456}}\\\underline{{\left. 2 \right|1728}}\\\underline{{\left. 2 \right|864}}\\\underline{{\left. 2 \right|432}}\\\underline{{\left. 2 \right|216}}\\\underline{{\left. 2 \right|108}}\\\underline{{\left. 2 \right|54}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 110592 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
So, cube root of 110592 = 2 × 2 × 2 × 2 × 3
= 48

(viii) 46656
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|46656}}\\\underline{{\left. 2 \right|23328}}\\\underline{{\left. 2 \right|11664}}\\\underline{{\left. 2 \right|5832}}\\\underline{{\left. 2 \right|2916}}\\\underline{{\left. 2 \right|1458}}\\\underline{{\left. 3 \right|729}}\\\underline{{\left. 3 \right|243}}\\\underline{{\left. 3 \right|81}}\\\underline{{\left. 3 \right|27}}\\\underline{{\left. 3 \right|9}}\\\underline{{\left. 3 \right|3}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
So, cube root of 46656 = 2 × 2 × 3 × 3
= 36

(ix) 175616
\(\displaystyle \begin{array}{l}\underline{{\left. 2 \right|175616}}\\\underline{{\left. 2 \right|87808}}\\\underline{{\left. 2 \right|43904}}\\\underline{{\left. 2 \right|21952}}\\\underline{{\left. 2 \right|10976}}\\\underline{{\left. 2 \right|5488}}\\\underline{{\left. 2 \right|2744}}\\\underline{{\left. 2 \right|1372}}\\\underline{{\left. 2 \right|686}}\\\underline{{\left. 7 \right|343}}\\\underline{{\left. 7 \right|49}}\\\underline{{\left. 7 \right|7}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7
So, cube root of 175616 = 2 × 2 × 2 × 7
= 56

(x) 91125
\(\displaystyle \begin{array}{l}\underline{{\left. 3 \right|91125}}\\\underline{{\left. 3 \right|30375}}\\\underline{{\left. 3 \right|10125}}\\\underline{{\left. 3 \right|3375}}\\\underline{{\left. 3 \right|1125}}\\\underline{{\left. 3 \right|375}}\\\underline{{\left. 5 \right|125}}\\\underline{{\left. 5 \right|25}}\\\underline{{\left. 5 \right|5}}\\\left. {\,\,} \right|1\end{array}\)
Prime factor of 91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
So, cube root of 91125 = 3 × 3 × 5
= 45

2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If 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.

Solution :
(i) Cube of any odd number is even.
False
 
(ii) A perfect cube does not end with two zeros.
True

(iii) If square of a number ends with 5, then its cube ends with 25.
False

(iv) There is no perfect cube which ends with 8.
False

(v) The cube of a two digit number may be a three digit number.
False

(vi) The cube of a two digit number may have seven or more digits.
False

(vii) The cube of a single digit number may be a single digit number.
True

3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution :
First, make groups of three digits, starting with the digit on the far right of 1331. The first group is 1331, the second group is 1 at the units place of 1 331, we know that if 1 is located at the end of a whole number, then it will have 1 at the units place of its cube root. So we get the unit’s digit of the cube root as 1, taking the second group 1, the cube of 1 corresponds to the number in the second group. Therefore, the ten’s digit of the cube root will be equal to the number in the units place of the smaller number, which is closest to the number in the second group. For example, 1, so 1 will be in the tens place of the cube root of 1331.
= \(\displaystyle \sqrt[3]{{1331}}\)
= 11
Similarly, the cube root of 4913 can be found, dividing it into two groups, the first group is 913, the second group 4 comes at the end of 913. When 3 comes at the end of a whole number, only 7 will come in the unit’s place of its cube root. So we take 7 in the unit’s place. Taking the second group, since 1<4<8
so
1³ = 1
2³ = 8
Therefore 1 will be taken as the tens digit of the cube root.
= \(\displaystyle \sqrt[3]{{4913}}\)
= 17