NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots

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Updated at 2021-02-11


NCERT solutions for class 8 Maths Chapter 7 Cube And Cube Roots Exercise 7.1

1.Which of the following numbers are not perfect cubes?
(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

(i) We have the prime factors of 216 as:

2216210825432739331

216=2×2×2×3×3×3

Thus, 216 is a perfect cube because in the factorisation, 2 and 3 have formed a group of three.

(ii) We have the prime factors of 128 as:

21282642322162824221

128=2×2×2×2×2×2×2

Here, 2 is left without making a group of three.So, 128 is not a perfect cube.

(iii) We have the prime factors of 1000 as:

21000250022505125525551

1000=2×2×2×5×5×5

Thus, 1000 is a perfect cube because all the factors make perfect pairs.

(iv) We have the prime factors of 100 as:

2100250525551

100=2×2×5×5

Thus, 100 is not a perfect cube because 2 and 5 have not formed a required pair of three.

(v) We have the prime factors of 46656 as:

2466562233282116642583222916214583729324338132739331

46656=2×2×2×2×2×2×3×3×3×3×3×3

Thus, 46656 is a perfect cube because all the factors are in perfect pairs of three.

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



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

(i) We have the prime factors of 243 as:

324338132739331

243=3×3×3×3×3=33×3×3

We can observe clearly that, number 3 is required to make 3×3 a group of three. Thus, the required smallest number to be multiplied is 3.

(ii) We have the prime factors of 256 as:

225621282642322162824221

256=2×2×2×2×2×2×2×2=23×23×2×2

We can observe clearly that, number 2 is required to make 2×2 a group of three.Thus, the required smallest number to be multiplied is 2.

(iii) We have the prime factors of 72 as:

27223621839331

72=2×2×2×3×3=23×3×3

We can observe clearly that, number 3 is required to make 3×3 a group of three.Thus, the required smallest number to be multiplied is 3.

(iv) We have the prime factors of 675 as:

36753225375525551

675=3×3×3×5×5=33×5×5

We can observe clearly that, number 5 is required to make 5×5 a group of three.Thus, the required smallest number to be multiplied is 5.

(v) We have the prime factors of 100 as:

2100250525551

243=2×2×5×5

We can observe clearly that, number 2 and 5 are required to make 2×2 and 5×5 a group of three.Thus, the required smallest number to be multiplied is 5×2=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) 92

(v) 704



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

(i) We have the prime factors of 81 as:

38132739331

81=3×3×3×3=33×3

We can observe clearly that, number 3 is divided to 81 to make it a perfect cube.

So, 81 ÷ 3 = 27 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(ii) We have the prime factors of 256 as:

21282642322162824221

128=2×2×2×2×2×2×2=23×23×2

We can observe clearly that, number 2 is divided to 128 to make it a perfect cube.

So, 128 ÷ 2 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 2.

(iii) We have the prime factors of 135 as:

3135345315551

81=3×3×3×5=33×5

We can observe clearly that, number 5 is divided to 135 to make it a perfect cube.

So, 135 ÷ 5= 27 which is a perfect cube. Thus, the required smallest number to be divided is 5.

(iv) We have the prime factors of 192 as:

219229624822421226331

192=2×2×2×2×2×2×3=23×23×3

We can observe clearly that, number 3 is divided to 192 to make it a perfect cube.

So, 192 ÷ 3 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 3.

(v) We have the prime factors of 704 as:

27042352217628824422211111

128=2×2×2×2×2×2×11=23×23×11

We can observe clearly that, number 11 is divided to 704 to make it a perfect cube.

So, 704 ÷ 11 = 64 which is a perfect cube. Thus, the required smallest number to be divided is 11.

4.Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

Given, the sides of the cuboid as 5 cm, 2 cm and 5 cm.

So we have the volume of the cuboid =5cm×2cm×5cm=50cm3

We have the prime factors of 50 as:

50=2×5×5

To make it a perfect cube, we must have 2×2×2×5×5×5

?20×(2×5×5)

?20×volumeofthegivencuboid

Thus,we have the required number of cuboids = 20.

NCERT solutions for class 8 Maths Chapter 7 Cube And Cube Roots 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



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

(i) We have the prime factors of 64 as:

2642322162824221

256=2×2×2×2×2×2=23×23

?643=2×2=4

Hence, the cube root of 64 is 4.

(ii) We have the prime factors of 512 as:

2512225621282642322162824221

256=2×2×2×2×2×2×2×2×2=23×23×23

?5123=2×2×2=8

Hence, the cube root of 512 is 8.

(iii) We have the prime factors of 10648 as:

21064825324226621113311112111111

10648=2×2×2×11×11×11=23×113×

?106483=2×11=22

Hence, the cube root of 10648 is 22.

(iv) We have the prime factors of 27000 as:

22700021350026750333753112533755125525551

27000=2×2×2×3×3×3×5×5×5=23×33×53

?270003=2×3×5=30

Hence, the cube root of 27000 is 30.

(v) We have the prime factors of 15625 as:

5156255312556255125525551

15625=5×5×5×5×5×5=53×53

?156253=5×5=25

Hence, the cube root of 15625 is 25.

(vi) We have the prime factors of 13824 as:

213824269122345621728286424322216210825432739331

13824=2×2×2×2×2×2×2×2×2×3×3×3

?23×23×23×33

?138243=2×2×2×3=24

Hence, the cube root of 13824 is 24.

(vii) We have the prime factors of 110592 as:

2110592255296227648213824269122345621728286424322216210825432739331

110592=2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

?23×23×23×23×33

?1105923=2×2×2×2×3=48

Hence, the cube root of 110592 is 48.

(viii) We have the prime factors of 46656 as:

2466562233282116642583222916214583729324338132739331

46656=2×2×2×2×2×2×3×3×3×3×3×3

?23×23×33×33

?466563=2×2×3×3=36

Hence, the cube root of 46656 is 36.

(ix) We have the prime factors of 175616 as:

217561628780824390422195221097625488227442137226867343749771

175616=2×2×2×2×2×2×2×2×2×7×7×7

?23×23×23×73

?1756163=2×2×2×7=56

Hence, the cube root of 175616 is 56.

(x) We have the prime factors of 91125 as:

391125330375310125333753112533755125525551

91125=3×3×3×3×3×3×5×5×5=33×33×53

?911253=3×3×5=45

Hence, the cube root of 91125 is 45.

2.State True or False.
(i) Cube of an 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.



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

(i) False because Cube of any odd number is always odd, e.g., (3)3=27

(ii) True.

(iii) True. For e.g., (5)2=25 and (5)3=625

(iv) False because there exists perfect cube which ends with 8 for e.g., (12)3=1728ending with 8

(v) False. let's take the lowest two digit number i.e., (10)3=1000which is a 4-digit number.

(vi) False.Let's take the highest two digit number i.e., (99)3=970299 which is 6-digit number.

(vii) True. For e.g., (2)3=8 which is 1-digit number.

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.



NCERT Solutions for Class 8 Maths Chapter 7 Cube And Cube Roots


Answer :

We have the perfect cube = 1331
Now let us form groups of three from the rightmost digits of 133

So we have,

Group 2= 1 and, group 1= 331

One’s digit in first group = 1, that says that one’s digit in the required cube root may be 1.Also, in the second group we have only 1.

Estimated cube root of 1331 = 11

Thus, 13313=11

(i) We have the perfect cube = 4913

Now let us form groups of three from the rightmost digits of 4913

So we have,

group 2 = 4 And,group 1= 913

One’s place digit in 913 is 3.So that says that one’s place digit in the cube root of the given number may be 7.

Now in group 2 digit is 4 and we know that 4 lies as:13<4<23.We know that, ten’s place must be the smallest number i.e,1.

Thus, the estimated cube root of 4913 = 17.

(ii) We have the perfect cube = 12167

Now let us form groups of three from the rightmost digits of 12167

So we have,

group 2= 12 and, group 1= 167

The ones place digit in 167 is 7.So we can say that one’s digit in the cube root of the given number may be 3.

Now in group 2, we have 12 and 12 lies between as:23<12<33. Also, we know that the ten’s place of the required cube root of the given number = 2

. Thus, the estimated cube root of 12167 = 23.

(iii) We have the perfect cube = 32768

Now let us form groups of three from the rightmost digits of 32768, we have

So we have,

group 2= 32 and, group 1= 768

One’s place digit in 768 is 8.So we can say that the one’s place digit in the cube root of the given number may be 2.

Now in group 2, we have 32 and 32 lies between the numbers as:33<32<43.Also, we know that the ten’s place of the cube root of the given number = 3.

Thus, the estimated cube root of 32768 = 32.



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