Cubes and Cube Roots-NCERT Solutions

Class VIII Math
NCERT Solution for Cubes and Cube Roots

Exercise 7.1 (Page 114)

Q1. Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Sol. (i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3

Grouping the prime factors of 216 into triples, no factor is left over.

∴ 216 is a perfect cube.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.

∴ 128 is not a perfect cube.

(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5

Grouping the prime factors of 1000 into triples, we are not left over with any factor.

∴ 1000 is a perfect cube.

(iv) We have 100 = 2 × 2 × 5 × 5

Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and 5 × 5 are not in triples.

∴ 100 is not a perfect cube.

(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Grouping the prime factors of 46656 in triples we are not left over with any prime factor.

∴ 46656 is a perfect cube.

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

Sol. (i) We have 243 = 3 × 3 × 3 × 3 × 3

The prime factor 3 is not a group of three.

∴ 243 is not a perfect cube.

Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3

or 729 =3 × 3 × 3 × 3 × 3 × 3

Now, 729 becomes a perfect cube.

Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 256 in triples, we are left over with 2 × 2.

∴ 256 is not a perfect cube.

Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2

or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

i.e. 512 is a perfect cube.

Thus, the required smallest number is 2.

(iii) We have 72 = 2 × 2 × 2 × 3 × 3

Grouping the prime factors of 72 in triples, we are left over with 3 × 3.

∴ 72 is not a perfect cube.

Now, [72] × 3 =[2 × 2 × 2 × 3 × 3] × 3

or 216 = 2 × 2 × 2 × 3 × 3 × 3

i.e. 216 is a perfect cube.

∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 × 3 × 3 × 5 × 5

Grouping the prime factors of 675 to triples, we are left over with 5 × 5.

∴ 675 is not a perfect cube.

Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5

or 3375 = 3 × 3 × 3 × 5 × 5 × 5

Now, 3375 is a perfect cube.

Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.

(v) We have 100 = 2 × 2 × 5 × 5

The prime factor are not in the groups of triples.

∴100 is not a perfect cube.

Now [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5

or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5

1000 = 2 × 2 × 2 × 5 × 5 × 5

Now, 1000 is a perfect cube.

Thus, the required smallest number is 10.

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

Sol. (i) We have 81 = 3 × 3 × 3 × 3

Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is not a perfect cube.

Now, [81] 3 = [3 × 3 × 3 × 3] + 3

or 27 = 3 × 3 × 3

i.e. 27 is a prefect cube

Thus, the required smallest number is 3.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left with 2.

∴ 128 is not a perfect cube

Now, [128] 2 = [2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

∴ The smallest required number is 2.

(iii) We have 135 = 3 × 3 × 3 × 5

Grouping the prime factors of 135 into triples, we are left over with 5.

∴ 135 is not a perfect cube

Now, [l35] 5 = [3 × 3 × 3 × 5] 5

or 27 = 3 × 3 × 3

i.e. 27 is a perfect cube.

Thus, the required smallest number is 5.

(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Grouping the prime factors of 192 into triples, 3 is left over.

∴ 192 is not a perfect cube.

Now, [192] 3 =[2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 3.

(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Grouping the prime factors of 704 into triples, 11 is left over.

∴ [704] 11 =[2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 11.

Q4. Parikshit makes a cuboid of plasticine of sidec 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Sol: Sides of the cuboid arc: 5 cm, 2 cm, 5 cm

∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm

To form it as a cube its dimension should be in the group of triples.

∴ Volume of the required cube = [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm

=[5 × 5 × 2 cm3] = 20 cm3

Thus, the required number of cuboids = 20.

Exercise 7.2 (page 116)

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

Sol. (i) By prime factorisation, we have

(ii) By prime factorisation, we have

(iii) By prime factorisation, we have

Thus, cube root of 10648 is 22.

(iv) By prime factorisation, we have

Thus, cube root of 27000 is 30.

(v) By prime factorisation, we have

Thus, cube root of 15625 is 25.

(vi) By prime factorisation, we have

Thus, cube root of 13824 is 24.

(vii) By prime factorisation, we have

Thus, the cube root of 110592 is 48

(viii) By the prime factorisation, we have

Thus, the cube root of 46656 is 32

(ix) By prime factorisation, we have

175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7

= 2 × 2 × 2 × 7 = 56

Thus the cube root of 175616 is 56.

(x) By prime factorisation, we have:

91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5

= 3 × 3 × 5 = 45

Thus, the cube root of 91125 is 45.

Q2. 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 he 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.

Sol. (i) False (ii) True (iii) False (iv) False

(v) False (vi) False (vii) True

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.

Sol. (i) Separating the given number (1331) into two groups:

1331 → 1 and 331

∴ 331 end in 1.

∴ Unit's digit of the cube root = 1

∴ Ten's digit of the cube root = I

(ii) Separating the given number (4913) in two groups:

4913 → 4 and 913

Unit's digits:

∵ Unit's digit in 913 is 3.

∴ Unit's digit of the cube root = 7

[7³ = 343; which ends in 3]

Ten's digit:

1³ = 1, 2³ = 8

and 1 < 4 < 8

i.e. 13 < 4 < 23

∴ The ten's digit of the cube root is 1.

(iii) Separating 12,167 in two groups:

12167 → 12 and 167

Unit's digit:

∵ 167 is ending in 7 and cube of a number ending in 3 ends in 7.

∴ The unit's digit of the cube root = 3

Ten's digit:

∵ 2³ = 8 and 3³ = 27

Also, 8 < 12 < 27

or 23 < 12 < 32

∴ The tens digit of the cube root can be 2.

(iv) Separating 32768 in two groups:

32768 → 32 and 786

Unit's digit:

768 will guess the unit's digit in the cube root.

∵ 768 ends in 8.

Unit's digit in the cube root = 2

Ten's digit:

∵ 3³ = 27 and 4³ – 64

Also, 27 < 32 < 64

or 33 < 32 < 43

The ten's digit of the cube root = 3.