#### NCERT Solutions for Class 8th: Ch 7 Cube and Cube Roots Maths

Page No: 114

Exercise 7.1

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

Solution

By resolving 216 into prime factor, 216 = 2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
216 = (2×2×2)×(3×3×3)
Here, 216 can be grouped into triplets of equal factors,
∴ 216 = (2×3) = 6

Hence, 216 is cube of 6.

(ii)  128

Solution

By resolving 128 into prime factor, 128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
128 = (2×2×2)×(2×2×2)×2
Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .
∴ 128 is not a perfect cube.

(iii) 1000

Solution

By resolving 1000 into prime factor, 1000 = 2×2×2×5×5×5
By grouping the factors in triplets of equal factors,
1000 = (2×2×2)×(5×5×5)
Here, 1000 can be grouped into triplets of equal factors,
∴ 1000 = (2×5) = 10
Hence, 1000 is cube of 10.

(iv) 100

Solution

By resolving 100 into prime factor, 100 = 2×2×5×5
Here, 100 cannot be grouped into triplets of equal factors.
∴ 100 is not a perfect cube.

(v)  46656

Solution

By resolving 46656 into prime factor, 46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 46656 is cube of 36.

2.Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

Solution

By resolving 243 into prime factor, 243 = 3×3×3×3×3
By grouping the factors in triplets of equal factors,
243 = (3×3×3)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 243 by 3 to get perfect square.

(ii) 256

Soliution

By resolving 256 into prime factor, 256 = 2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
256 = (2×2×2)×(2×2×2)×2×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will multiply 256 by 2 to get perfect square.

(iii) 72

Solution

By resolving 72 into prime factor, 72 = 2×2×2×3×3
By grouping the factors in triplets of equal factors,
72 = (2×2×2)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 72 by 3 to get perfect square.

(iv)  675

Solution

By resolving 675 into prime factor, 675 = 3×3×3×5×5
By grouping the factors in triplets of equal factors,
675 = (3×3×3)×5×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 675 by 5 to get perfect square.

(v) 100

Solution

By resolving 100 into prime factor, 100 = 2×2×5×5
Here, 2 and 5  cannot be grouped into triplets of equal factors.
∴ We will multiply 100 by (2×5) 10 to get perfect square.

3.Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

Solution

By resolving 81 into prime factor, 81 = 3×3×3×3
By grouping the factors in triplets of equal factors,
81 = (3×3×3)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 81 by 3 to get perfect square.

(ii) 128

Solution

By resolving 128 into prime factor, 128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
128 = (2×2×2)×(2×2×2)×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will divide 128 by 2 to get perfect square.

(iii) 135

Solution

By resolving 135 into prime factor, 135 = 3×3×3×5
By grouping the factors in triplets of equal factors,
135 = (3×3×3)×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will divide 135 by 5 to get perfect square.

(iv) 192

Solution

By resolving 192 into prime factor, 192 = 2×2×2×2×2×2×3
By grouping the factors in triplets of equal factors,
192 = (2×2×2)×(2×2×2)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 192 by 3 to get perfect square.

(v)  704

Solution

By resolving 704 into prime factor, 704 = 2×2×2×2×2×2×11
By grouping the factors in triplets of equal factors,
704 = (2×2×2)×(2×2×2)×11
Here, 11 cannot be grouped into triplets of equal factors.
∴ We will divide 704 by 11 to get perfect square.

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

Given, side of cube is 5 cm, 2 cm and 5 cm.
∴ Volume of cube = 5×2×5 = 50 50 = 2×5×5
Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 50 by (2×2×5) 20  to get perfect square.
Hence, 20 cuboid is needed.

Page No. 116

Exercise 7.2

1. Find the cube root of each of the following numbers by prime factorisation method
(i) 64

Solution

64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.

(ii) 512

Solution

512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.

(iii) 10648

Solution

10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors,
10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.

(iv) 27000

Solution

27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors,
27000 = (2×2×2)×(3×3×3)×(5×5×5)
Here, 27000 can be grouped into triplets of equal factors,
∴ 27000 = (2×3×5) = 30
Hence, 30 is cube root of 27000.

(v) 15625

Solution

15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors,
15625 = (5×5×5)×(5×5×5)
Here, 15625 can be grouped into triplets of equal factors,
∴ 15625 = (5×5) = 25
Hence, 25 is cube root of 15625.

(vi) 13824

Solution

13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 13824 can be grouped into triplets of equal factors,
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is cube root of 13824.

(vii) 110592

Solution

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 110592 can be grouped into triplets of equal factors,
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is cube root of 110592.

(viii) 46656

Solution

46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 36 is cube root of 46656.

(ix) 175616

Solution

175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)
Here, 175616 can be grouped into triplets of equal factors,
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is cube root of 175616.

(x) 91125

Solution

91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors,
91125 = (3×3×3)×(3×3×3)×(5×5×5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3×3×5) = 45
Hence, 45 is cube root of 91125.

2.State true or false.
(i) Cube of any odd number is even.

Ans.False

(ii) A perfect cube does not end with two zeros.

Ans.True

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

Ans.False

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

Ans.False

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

Ans.False

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

Ans.False

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

Ans.True

4. 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

• By grouping the digits, we get 1 and 331
As we know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ We get 1 as unit digit of the cube root of 1331.
The cube of 1 matches with the number of second group.
∴ The ten digit of our cube root is taken as the unit place of smallest number.
As we know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ = 11

• By grouping the digits, we get 4 and 913
As we know that the unit’s digit of the cube of a number having digit as unit’s place 3 is 7.
∴ we get 7 as unit digit of the cube root of 4913.
We know 1= 1  and 2 = 8 , 1 > 4 > 8.
Thus, 1 is taken as ten digit of  cube root.
∴ = 17

• By grouping the digits, we get 12 and 167.
As we know that the unit’s digit of the cube of a number having digit as unit’s place 7 is 3.
∴ we get 3 as unit digit of the cube root of 12167
We know 2= 8  and 3 = 27 , 8 > 12 > 27.
Thus, 2 is taken as ten digit of  cube root.
∴ = 23

• By grouping the digits, we get 32 and 768.
As we know that the unit’s digit of the cube of a number having digit as unit’s place 8 is 2.
∴ we get 2 as unit digit of the cube root of 32768.
We know 33 = 27  and 43 = 64 , 27 > 32 > 64.
Thus,  3 is taken as ten digit of  cube root.
∴ = 32

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