NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots
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Page 1
NCERT
SOLUTIONS
CLASS - 8TH
aglase .co
SOLUTIONS
CLASS - 8TH
aglase .co
Page 2
Book : Mathematics Ncert Solutions | Chapter-7 Maths
Class : 8th
Subject : Maths
Chapter : 7
Chapter Name : Cubes and Cube Roots
Exercise 7.1
Q1 Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Answer. (i) The prime factorisation of 216 is as follows:
3 3
216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 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:
Page 1 of 8 Aglasem Schools
Class : 8th
Subject : Maths
Chapter : 7
Chapter Name : Cubes and Cube Roots
Exercise 7.1
Q1 Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656
Answer. (i) The prime factorisation of 216 is as follows:
3 3
216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 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:
Page 1 of 8 Aglasem Schools
Page 3
Book : Mathematics Ncert Solutions | Chapter-7 Maths
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:
Page 2 of 8 Aglasem Schools
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:
Page 2 of 8 Aglasem Schools
Page 4
Book : Mathematics Ncert Solutions | Chapter-7 Maths
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.
Page : 114 , Block Name : Exercise 7.1
Q2 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
Answer. (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
Page 3 of 8 Aglasem Schools
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.
Page : 114 , Block Name : Exercise 7.1
Q2 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
Answer. (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
Page 3 of 8 Aglasem Schools
Page 5
Book : Mathematics Ncert Solutions | Chapter-7 Maths
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
Page : 114 , Block Name : Exercise 7.1
Q3 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
Answer. (i) 81 = 3 x 3 x 3 x 3
Here, one 3 is left which is not in a triplet.
If we divide 81 by 3, then it will become a perfect cube.
Thus, 81 3 = 27 = 3 x 3 x 3 is a perfect cube.
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.
(ii) 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, one 2 is left which is not in a triplet.
If we divide 128 by 2, then it will become a perfect cube.
Thus, 128 ÷ 2 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.
(iii) 135 = 3 x 3 x 3 x 5.
Here, one 5 is left which is not in a triplet.
If we divide 135 by 5, then it will become a perfect cube.
Thus, 135 ÷ 5 = 27 = 3 x 3 x 3 is a perfect cube.
Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.
(iv) 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3.
Page 4 of 8 Aglasem Schools
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
Page : 114 , Block Name : Exercise 7.1
Q3 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
Answer. (i) 81 = 3 x 3 x 3 x 3
Here, one 3 is left which is not in a triplet.
If we divide 81 by 3, then it will become a perfect cube.
Thus, 81 3 = 27 = 3 x 3 x 3 is a perfect cube.
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.
(ii) 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Here, one 2 is left which is not in a triplet.
If we divide 128 by 2, then it will become a perfect cube.
Thus, 128 ÷ 2 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.
(iii) 135 = 3 x 3 x 3 x 5.
Here, one 5 is left which is not in a triplet.
If we divide 135 by 5, then it will become a perfect cube.
Thus, 135 ÷ 5 = 27 = 3 x 3 x 3 is a perfect cube.
Hence, the smallest number by which 135 should be divided to make it a perfect cube is 5.
(iv) 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3.
Page 4 of 8 Aglasem Schools
Page 6
Book : Mathematics Ncert Solutions | Chapter-7 Maths
Here, one 3 is left which is not in a triplet.
If we divide 192 by 3, then it will become a perfect cube.
Thus, 192 ÷ 3 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.
(v) 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11.
Here, one 11 is left which is not in a triplet.
If we divide 704 by 11, then it will become a perfect cube.
Thus, 704 ÷ 11 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.
Page : 114 , Block Name : Exercise 7.1
Q4 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?
Answer. Here, some cuboids of size 5 x 2 x 5 are given.
When these cuboids are arranged to form a cube, the side of this cube so formed will be a
common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.
LCM of 5, 2, and 5 = 10.
Let us try to make a cube of 10 cm side.
For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width,
and 2 along with its height.
Total cuboids required according to this arrangement = 2 x 5 x 2 = 20.
With the help of 20 cuboids of such measures, a cube is formed as follows.
Alternatively
Volume of the cube of sides 5 cm, 2 cm, 5 cm
= 5 cm x 2 cm x 5 cm = (5 x 5 x 2) cm³.
Here, two 5s and one 2 are left which are not in a triplet.
If we multiply this expression by 2 x 2 x 5 20, then it will become a perfect cube.
Thus, (5 x 5 x 2 x 2 x 2 x 5) = x = ( 5 x 5 x 5 x 2 x 2 x 2 ) = 1000 is a perfect cube.
Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.
Page : 114 , Block Name : Exercise 7.1
Page 5 of 8 Aglasem Schools
Here, one 3 is left which is not in a triplet.
If we divide 192 by 3, then it will become a perfect cube.
Thus, 192 ÷ 3 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 192 should be divided to make it a perfect cube is 3.
(v) 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11.
Here, one 11 is left which is not in a triplet.
If we divide 704 by 11, then it will become a perfect cube.
Thus, 704 ÷ 11 = 64 = 2 x 2 x 2 x 2 x 2 x 2 is a perfect cube.
Hence, the smallest number by which 704 should be divided to make it a perfect cube is 11.
Page : 114 , Block Name : Exercise 7.1
Q4 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?
Answer. Here, some cuboids of size 5 x 2 x 5 are given.
When these cuboids are arranged to form a cube, the side of this cube so formed will be a
common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.
LCM of 5, 2, and 5 = 10.
Let us try to make a cube of 10 cm side.
For this arrangement, we have to put 2 cuboids along with its length, 5 along with its width,
and 2 along with its height.
Total cuboids required according to this arrangement = 2 x 5 x 2 = 20.
With the help of 20 cuboids of such measures, a cube is formed as follows.
Alternatively
Volume of the cube of sides 5 cm, 2 cm, 5 cm
= 5 cm x 2 cm x 5 cm = (5 x 5 x 2) cm³.
Here, two 5s and one 2 are left which are not in a triplet.
If we multiply this expression by 2 x 2 x 5 20, then it will become a perfect cube.
Thus, (5 x 5 x 2 x 2 x 2 x 5) = x = ( 5 x 5 x 5 x 2 x 2 x 2 ) = 1000 is a perfect cube.
Hence, 20 cuboids of 5 cm, 2 cm, 5 cm are required to form a cube.
Page : 114 , Block Name : Exercise 7.1
Page 5 of 8 Aglasem Schools
Page 7
Book : Mathematics Ncert Solutions | Chapter-7 Maths
Exercise 7.2
Q1 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
Answer. (i) Prime factorisation of 64 = 2 x 2 x 2 x 2 x 2 x 2
3
∴ √64 = 2 × 2 = 4
(ii) Prime factorisation of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
3
∴ √512 = 2 × 2 × 2 = 8
(iii) Prime factorisation of 10648 = 2 × 2 × 2 × 11 × 11 × 11
3
∴ √10648 = 2 × 11 = 22
(iv) Prime factorisation of 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
3
∴ √27000 = 2 × 3 × 5 = 30
Prime factorisation of 15625 = 5 × 5 × 5 × 5 × 5 × 5
(v) 3
√15625 = 5 × 5 = 25
(vi) Prime factorisation of 13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
3
∴ √13824 = 2 × 2 × 2 × 3 = 24
(vii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(viii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(viii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(ix) Prime factorisation of 175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7
3
∴ √175616 = 2 × 2 × 2 × 7 = 56
(x) Prime factorisation of 91125 = 3 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5
3
∴ √91125 = 3 × 3 × 5 = 45
Page : 116 , Block Name : Exercise 7.2
Q2 State true or false.
(i) Cube of any odd number is even.
Page 6 of 8 Aglasem Schools
Exercise 7.2
Q1 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
Answer. (i) Prime factorisation of 64 = 2 x 2 x 2 x 2 x 2 x 2
3
∴ √64 = 2 × 2 = 4
(ii) Prime factorisation of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
3
∴ √512 = 2 × 2 × 2 = 8
(iii) Prime factorisation of 10648 = 2 × 2 × 2 × 11 × 11 × 11
3
∴ √10648 = 2 × 11 = 22
(iv) Prime factorisation of 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
3
∴ √27000 = 2 × 3 × 5 = 30
Prime factorisation of 15625 = 5 × 5 × 5 × 5 × 5 × 5
(v) 3
√15625 = 5 × 5 = 25
(vi) Prime factorisation of 13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
3
∴ √13824 = 2 × 2 × 2 × 3 = 24
(vii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(viii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(viii) Prime factorisation of 46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3
3
∴ √46656 = 2 × 2 × 3 × 3 = 36
(ix) Prime factorisation of 175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7
3
∴ √175616 = 2 × 2 × 2 × 7 = 56
(x) Prime factorisation of 91125 = 3 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5
3
∴ √91125 = 3 × 3 × 5 = 45
Page : 116 , Block Name : Exercise 7.2
Q2 State true or false.
(i) Cube of any odd number is even.
Page 6 of 8 Aglasem Schools
Page 8
Book : Mathematics Ncert Solutions | Chapter-7 Maths
(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.
Answer. For nding the cube of any number, the number is rst multiplied with itself and this
product is again multiplied with this number.
(i) False. When we nd out the cube of an odd number, we will nd an odd number as the
result because the unit place digit of an odd number is odd and we are multiplying three odd
numbers. Therefore, the product will be again an odd number. For example, the cube of 3 (i.e.,
an odd number) is 27, which is again an odd number.
(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect
multiple of 3.
For example, the cube of 10 is 1000 and there are 3 zeroes at the end of it.
The cube of 100 is 1000000 and there are 6 zeroes at the end of it.
(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube
will end with 25.
For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625.
However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is
42875 which does not end with 25.
(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having
their unit place digit as 2 will end with 8. The cube of 12 is 1728 and the cube of 22 is 10648.
(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4
digits in it.
(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6
digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.
(vii) True, as the cube of 1 and 2 are 1 and 8 respectively.
Page : 116 , Block Name : Exercise 7.2
Q3 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.
Answer. Firstly, we will make groups of three digits starting from the rightmost digit of the
number as 1331.
There are 2 groups, 1 and 331, in it.
Considering 331,
The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube
number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place
digit of the required cube root can be taken as 1.
Taking the other group i.e., 1,
The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit
of our cube root will be taken as the unit place of the smaller number whose cube is near to the
Page 7 of 8 Aglasem Schools
(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.
Answer. For nding the cube of any number, the number is rst multiplied with itself and this
product is again multiplied with this number.
(i) False. When we nd out the cube of an odd number, we will nd an odd number as the
result because the unit place digit of an odd number is odd and we are multiplying three odd
numbers. Therefore, the product will be again an odd number. For example, the cube of 3 (i.e.,
an odd number) is 27, which is again an odd number.
(ii) True. Perfect cube will end with a certain number of zeroes that are always a perfect
multiple of 3.
For example, the cube of 10 is 1000 and there are 3 zeroes at the end of it.
The cube of 100 is 1000000 and there are 6 zeroes at the end of it.
(iii) False. It is not always necessary that if the square of a number ends with 5, then its cube
will end with 25.
For example, the square of 25 is 625 and 625 has its unit digit as 5. The cube of 25 is 15625.
However, the square of 35 is 1225 and also has its unit place digit as 5 but the cube of 35 is
42875 which does not end with 25.
(iv) False. There are many cubes which will end with 8. The cubes of all the numbers having
their unit place digit as 2 will end with 8. The cube of 12 is 1728 and the cube of 22 is 10648.
(v) False. The smallest two-digit natural number is 10, and the cube of 10 is 1000 which has 4
digits in it.
(vi) False. The largest two-digit natural number is 99, and the cube of 99 is 970299 which has 6
digits in it. Therefore, the cube of any two-digit number cannot have 7 or more digits in it.
(vii) True, as the cube of 1 and 2 are 1 and 8 respectively.
Page : 116 , Block Name : Exercise 7.2
Q3 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.
Answer. Firstly, we will make groups of three digits starting from the rightmost digit of the
number as 1331.
There are 2 groups, 1 and 331, in it.
Considering 331,
The digit at its unit place is 1. We know that if the digit 1 is at the end of a perfect cube
number, then its cube root will have its unit place digit as 1 only. Therefore, the unit place
digit of the required cube root can be taken as 1.
Taking the other group i.e., 1,
The cube of 1 exactly matches with the number of the second group. Therefore, the tens digit
of our cube root will be taken as the unit place of the smaller number whose cube is near to the
Page 7 of 8 Aglasem Schools
Page 9
Book : Mathematics Ncert Solutions | Chapter-7 Maths
number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.
Hence, √1331 = 11
3
The cube root Of 4913 has to be calculated.
We will make groups of three digits starting from the rightmost digit of 4913, as 4913. The
groups are 4 and 913.
Considering the group 913,
The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube
number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place
digit of the required cube root is taken as 7.
Taking the other group i.e., 4,
We know that 1 = 1 and 2 = 8
3 3
Also, 1 < 4 < 8
Therefore, 1 will be taken at the tens place of the required cube root.
Thus, = √4913 = 17
3
The cube root Of 12167 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 12167 as
12167. The groups are 12 and 167.
12167, as 12167.
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
Considering the group 167.
167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its
cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required
cube root can be taken as 3.
Taking the other group i.e., 12,
We know that, 2 = 8 and 3 = 27
3 3
Also, 8 < 12 < 27,
2 is smaller between 2 and 3. Therefore, 2 will be taken at tens at the tens place of the required
cube root.
Thus, √12167 = 23.
3
The cube root of 32768 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 32768
32768, as 32768.
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
Considering the group 768,
768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its
cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required
cube root will be taken as 2.
Taking the other group i.e., 32,
We know that, 3 = 27 and 4 = 64
3 3
Also, 27 < 32 < 64
3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube
root.
Thus, √32768 = 32.
3
Page : 116 , Block Name : Exercise 7.2
Page 8 of 8 Aglasem Schools
number of the second group i.e., 1 itself. 1 will be taken as tens place of the cube root of 1331.
Hence, √1331 = 11
3
The cube root Of 4913 has to be calculated.
We will make groups of three digits starting from the rightmost digit of 4913, as 4913. The
groups are 4 and 913.
Considering the group 913,
The number 913 ends with 3. We know that if the digit 3 is at the end of a perfect cube
number, then its cube root will have its unit place digit as 7 only. Therefore, the unit place
digit of the required cube root is taken as 7.
Taking the other group i.e., 4,
We know that 1 = 1 and 2 = 8
3 3
Also, 1 < 4 < 8
Therefore, 1 will be taken at the tens place of the required cube root.
Thus, = √4913 = 17
3
The cube root Of 12167 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 12167 as
12167. The groups are 12 and 167.
12167, as 12167.
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
Considering the group 167.
167 ends with 7. We know that if the digit 7 is at the end of a perfect cube number, then its
cube root will have its unit place digit as 3 only. Therefore, the unit place digit of the required
cube root can be taken as 3.
Taking the other group i.e., 12,
We know that, 2 = 8 and 3 = 27
3 3
Also, 8 < 12 < 27,
2 is smaller between 2 and 3. Therefore, 2 will be taken at tens at the tens place of the required
cube root.
Thus, √12167 = 23.
3
The cube root of 32768 has to be calculated.
We will make groups of three digits starting from the rightmost digit of the number 32768
32768, as 32768.
¯
¯¯¯
¯¯¯¯
¯¯¯
¯¯¯
Considering the group 768,
768 ends with 8. We know that if the digit 8 is at the end of a perfect cube number, then its
cube root will have its unit place digit as 2 only. Therefore, the unit place digit of the required
cube root will be taken as 2.
Taking the other group i.e., 32,
We know that, 3 = 27 and 4 = 64
3 3
Also, 27 < 32 < 64
3 is smaller between 3 and 4. Therefore, 3 will be taken at the tens place of the required cube
root.
Thus, √32768 = 32.
3
Page : 116 , Block Name : Exercise 7.2
Page 8 of 8 Aglasem Schools
Document Details
| Board / Org | NCERT |
| Exam | Class 8 |
| Type | Solution |
| Pages | 9 |
| Language | english |
| Updated | 22 Jul 2026 |