7. Cubes and Cube Roots

Chapter 7 . Cubes and Cube Roots

7. Cubes 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

Solution: (i) 216

By prime factorisation ,we have

The prime factor 2 and 3 are appeared in a group of three . So, 216 is a perfect cube .

(ii) 128

By prime factorisation ,we have

Since there is only one 2 in the prime factorisation . So, we need (2 × 2 =) 4 to make it a perfect cube .

Therefore, 128 is not a perfect cube .

(iii) 1000

By prime factorisation ,we have

The prime factor 2 and 5 are appeared in a group of three . So, 1000 is a perfect cube .

(iv) 100

By prime factorisation ,we have

The prime factor 2 and 5 are not appeared in a group of three . So, 100 is not a perfect cube .

(v) 46656

By prime factorisation ,we have

The prime factor 2 and 3 are appeared in a group of three . So, 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

We have ,

In this prime factorisation ,we find that there is no triplet of 3 . So, 243 is not a perfect cube .

To make it a perfect cube we multiply it by 3 .

Thus , is a perfect cube .

(ii) 256

We have,

In this prime factorisation ,we find that there is no triplet of 2 . So, 256 is not a perfect cube .

To make it a perfect cube we multiply it by 2 .

Thus , is a perfect cube .

(iii) 72

We have ,

In this prime factorisation ,we find that there is no triplet of 2 . So, 256 is not a perfect cube .

To make it a perfect cube we multiply it by 2 .

Thus , is a perfect cube .

(iv) 675

We have ,

In this prime factorisation ,we find that there is no triplet of 5 . So, 675 is not a perfect cube .

To make it a perfect cube we multiply it by 5 .

Thus , is a perfect cube .

(v) 100

We have ,

In this prime factorisation ,we find that there is no triplet of 2 and 5 . So, 100 is not a perfect cube .

To make it a perfect cube we multiply it by 2 × 5 .

Thus , is a perfect cube .

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

We have ,

The primes 3 do not appear in groups of three . So, 81 is not a perfect cube .

If we divide 81 by 3 , then the prime factorization of the quotient will not contain 3 . So, the resulting perfect cube is .

(ii) 128

We have ,

The primes 2 do not appear in groups of three . So, 128 is not a perfect cube .

If we divide 128 by 2 , then the prime factorization of the quotient will not contain 2 . So, the resulting perfect cube is .

(iii) 135

We have ,

The primes 5 do not appear in groups of three . So, 135 is not a perfect cube .

If we divide 135 by 5 , then the prime factorization of the quotient will not contain 5 . So, the resulting perfect cube is .

(iv) 192

We have,

The primes 3 do not appear in groups of three . So, 192 is not a perfect cube .

If we divide 192 by 3 , then the prime factorization of the quotient will not contain 3 . So, the resulting perfect cube is .

(v) 704

We have ,

The primes 11 do not appear in groups of three . So, 81 is not a perfect cube .

If we divide 81 by 11 , then the prime factorization of the quotient will not contain 11 . So, the resulting perfect cube is .

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 , the sides of the cuboid are 5 cm , 2 cm and 5 cm .

We have ,

In this prime factorisation ,we find that there is no triplet of 2 and 5 . So, 50 is not a perfect cube .

To make it a perfect cube we multiply it by 2 × 2 × 5 .

Thus , is a perfect cube .

Hence , he needs 2×2×5 = 20 cuboids .

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

By prime factorization method , we have , 64 = 2×2×2×2×2×2

So,

(ii) 512

By prime factorization method , we have , 512 = 2×2×2×2×2×2×2×2×2

So,

(iii) 10648

By prime factorization method , we have,

10648 = 2 × 2 × 2 × 11 × 11 × 11

So,

(iv) 27000

By prime factorization method , we have 27000 = 2×2×2×3×3×3×5×5×5

So,

(v) 15625

By prime factorization method , we have 15625 = 5×5×5×5×5×5

So,

(vi) 13824

By prime factorization method , we have

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

So,

(vii) 110592

By prime factorization method , we have

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

So,

(viii) 46656

By prime factorization method , we have

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

So,

(ix) 175616

By prime factorization method ,we have

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

So,

(x) 91125

By prime factorization method , we have

91125 = 3×3×3×3×3×3×5×5×5

So,

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.

Answer: (i) False (ii) True (iii) False (iv) False (v) False (vi) False (vii) 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: (a) 1331

We have , 1331

We get 1 is second group and 331 is first groups as two groups of one digit and three digits each .

The number 331 ends with 1 , We know that 1 comes at the unit’s place of a number only when it’s cube root ends in 1 . So, we get 1 at the unit’s place of the cube root .

For second group , i.e. , 1

We know that 1³ = 1

Therefore , 1 will be taken at the tens place of the required cube root .

So , .

(b) 4913

We have , 4913

We get 4 is second group and 913 is first groups as two groups of one digit and three digits each .

The number 913 ends with 3 , We know that 7 comes at the unit’s place of a number only when it’s cube root ends in 3 . So, we get 7 at the unit’s place of the cube root .

For second group , i.e. , 4

We know that and

Also , 1 < 4 < 8

We take the one’s place of the smallest number 4 as the ten’s place of the required cube root . So, we get .

(b) 12167

We have , 12167

We get 12 is second group and 167 is first groups as two groups of one digit and three digits each .

The number 167 ends with 7 , We know that 3 comes at the unit’s place of a number only when it’s cube root ends in 7 . So, we get 3 at the unit’s place of the cube root .

For second group , i.e. , 12

We know that and

Also , 8 < 12 < 27

We take the one’s place of the smallest number 2 as the ten’s place of the required cube root . So, we get .

We have , 32768

We get 32 is second group and 768 is first groups as two groups of one digit and three digits each .

The number 768 ends with 2 , We know that 2 comes at the unit’s place of a number only when it’s cube root ends in 8 . So, we get 2 at the unit’s place of the cube root .

For second group , i.e. , 32

We know that and

Also , 27 < 32 < 64

We take the one’s place of the smallest number 3 as the ten’s place of the required cube root . So, we get .

Extra Questions :

(d) 592704

We have , 592704

We get 592 is second group and 704 is first groups as two groups of one digit and three digits each .

The number 704 ends with 4 , We know that 4 comes at the unit’s place of a number only when it’s cube root ends in 4 . So, we get 4 at the unit’s place of the cube root .

For second group , i.e. , 592

We know that and

Also , 512 < 592 < 729

We take the one’s place of the smallest number 8 as the ten’s place of the required cube root . So, we get .

(e) 912673

We have , 912673

We get 912 is second group and 673 is first groups as two groups of one digit and three digits each .

The number 673 ends with 7 , We know that 7 comes at the unit’s place of a number only when it’s cube root ends in 3 . So, we get 7 at the unit’s place of the cube root .

For second group , i.e. , 673

We know that and

Also , 512 < 673 < 729

We take the one’s place of the smallest number 8 as the ten’s place of the required cube root . So, we get .