Lesson: Cubes and Cube Roots

Exercise 7.1

Question 1

Which of the following numbers are not perfect cubes?

i) 216 (ii) 128 (iii) 1000

(iv) 100 (v) 46656

Answer:

(ii) and (iv)

Solution:

(i) 216 = 2 × 2 × 2 × 3 × 3 × 3

= 2

3

× 3

3

= (2 × 3)

3

= (6)

3

Hence, 216 is a perfect cube.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2

3

× 2

3

× 2

2 does not appear in a group of three.

Hence, 128 is not a perfect cube.

(iii) 1000 = 2 × 2 × 2 × 5 × 5 × 5

= 2

3

× 5

3

= 10

3

Hence, 1000 is perfect cube.

(iv) 100 = 2 × 2 × 5 × 5

2 and 5 both do not appear in a group of three.

Hence, 100 is not a perfect cube

(v) 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 9 × 9 × 9

= 2 × 2 × 9

= 36

Hence, 46656 is a perfect cube.

Question 2

Find the smallest number by which each of the following number must be

multiplied to obtain a perfect cube.

(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100

Answer:

(i) 3 (ii) 2 (iii) 3 (iv) 5 (v) 10

Solution:

(i) 243 = 3 × 3 × 3 × 3 × 3

The prime factor 3 does not appear in a group of three.

Therefore, 243 is not a perfect cube.

To make it a cube, we need one more 3.

In that case,

243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729, which is a perfect cube.

Hence, the required smallest number by which 243 should be multiplied to

make a perfect cube is 3.

(ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

The prime factor 2 does not appear in a group of three.

Therefore, 256 is not a perfect cube.

To make it a cube, we need one more 2.

In that case,

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512,

which is a perfect cube.

Hence, the required smallest number by which 256 should be multiplied to

make a perfect cube is 2.

(iii) 72 = 2 × 2 × 2 × 3 × 3

The prime factor 3 does not appear in a group of three.

Therefore, 72 is not a perfect cube.

To make it a cube, we need one more 3.

In that case,

72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216, which is a perfect cube.

Hence, the required smallest number by which 72 should be multiplied to

make a perfect cube is 3.

(iv) 675 = 3 × 3 × 3 × 5 × 5

The prime factor 5 does not appear in a group of three.

Therefore, 675 is not a perfect cube.

To make it a cube, we need one more 5.

In that case,

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375, which is a perfect cube.

Hence, the required smallest number by which 675 should be multiplied to

make a perfect cube is 5.

(v) 100 = 2 × 2 × 5 × 5

The prime factors 2 and 5 do not appear in a group of three.

Therefore, 100 is not a perfect cube.

To make it a cube, we need one more 2 and 5respectively.

In that case,

100 × 10 = 2 × 2 × 2 × 5 × 5 × 5 = 1000, which is a perfec cube.

Hence, the required smallest number by which 100 should be multiplied to

make a perfect cube is (2 × 5), i.e, 10.

Question 3

Find the smallest number by which each of the following number must be

divided to obtain a perfect cube.

(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704

Answer:

(i) 3 (ii) 2 (iii) 5 (iv) 3 (v) 11

Solutin:

(i) 81 = 3 × 3 × 3 × 3

The prime factor 3 does not appear in a group of three.

Therefore, 81 is not a perfect cube.

In the factorization, after a triplet of 3’s, one 3 is left alone..

So, if we divide 81 by 3, then the prime factorisation of the quotient will not

contain this 3.

81 ÷ 3 = 3 × 3 × 3

Further, the perfect cube in that case is 81 ÷ 3 = 27.

Hence, the smallest whole number by which 81 should be divided to make it a

perfect cube is 3.

(ii) 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

The prime factor 2 does not appear in a group of three.

Therefore, 128 is not a perfect cube.

In the factorization, after two triplets of 2’s, one

2 is left alone.

So, if we divide 128 by 2, then the prime factorisation of the quotient will not contain

this 2.

128 ÷ 2 = 2 × 2 × 2 × 2 × 2 × 2

Further, the perfect cube in that case is 128 ÷ 2 = 64.

Hence, the smallest whole number by which 128 should be divided to make it a

perfect cube is 2.

(iii) 135 = 3 × 3 × 3 × 5

The prime factor 5 does not appear in a group of three.

Therefore, 135 is not a

perfect cube. In the factorization, after a triplet of 3’s, one 5 is left alone.

So, if we divide 135 by 5, then the prime factorisation of the quotient will not

contain this5.

135 ÷ 5 = 3 × 3 × 3

Further, the perfect cube in that case is 135 ÷ 5 = 27

Hence, the smallest whole number by which 135 should be divided to make it a

perfect cube is 5.

(iv) 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3.

The prime factor 3 does not appear in a group of three.

Therefore, 192 is not a perfect cube.

In the factorization, after two triplets of 2’s, one 3 is left alone.

So, if we divide 192 by 3, then the prime factorisation of the quotient will not

contain this 3.

192 ÷ 3 = 2 × 2 × 2 × 2 × 2 × 2

Further, the perfect cube in that case is 192 ÷ 3 = 64

Hence, the smallest whole numbers by which 192 should be divided to make it

a perfect cube is 3.

(v) 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

The prime factor 11 does not appear in a group of three.

Therefore, 704 is not a perfect cube.

In the factorization, after two triplets of 2’s, one 11 is left alone.

So, if we divide 704 by 11, then the prime factorisation of the quotient will not

contain this 11.

704÷ 11 = 2 × 2 × 2 × 2 × 2 × 2

Further the perfect cube in that case is 704 ÷ 11 = 64.

Hence, the smallest whole number by which 704 should be divided to make it

aperfect cube is 11.

Question 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?

Answer:

20 cuboids

Solution:

The volume of Parikshit’s cuboid = l × b × h = 5 cm × 2 cm × 5 cm .

To make it a perfect cube, we need to multiply this by 5 × 2 × 2 = 20, so as to

make prime factors in groups of triples.

Hence, 20 such cubes will be required to make a cube.

The volume of this cube will be 10 cm × 10 cm × 10 cm = 1000 cm

3

.

Exercise 7.2

Question 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

Answer:

(i) 4 (ii) 8 (iii) 22 (iv) 30 (v) 25 (vi) 24

(vii) 48 (viii) 36 (ix) 56

Solution:

(i) 64 =

2 × 2 × 2 × 2 × 2 × 2

= 2

3

× 2

3

3

64

= 2 × 2 = 4

(ii) 512 =

2 × 2 × 2 × 2 × 2 × 2 2 × 2 × 2

= 2

3

×2

3

×2

3

3

512

= 2× 2 × 2 = 8

(iii) 10648 =

2 × 2 × 2 × 11 × 11 × 11

= 2

3

× 11

3

3

10648

= 2 × 11 = 22

(iv) 27000

= 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5

= 2

3

× 3

3

× 5

3

3

27000

= 2 × 3 × 5 = 30

(v) 15625 = 5 × 5 × 5 × 5 × 5 × 5

= 5

3

× 5

3

3

15625

= 5 × 5

3

15625

= 25

(vi) 13824 =

2 × 2 × 2 × 2 × 2 × 2

× 2 × 2 × 2 3 × 3 × 3

= 2

3

× 2

3

× 2

3

× 3

3

3

13824

= 2 × 2 × 2 × 3 = 24

(vii) 110592

2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 3 × 3 × 3

= 2

3

× 2

3

× 2

3

× 2

3

× 3

3

3

1100592

= 2 × 2 × 2 × 2 × 3 = 48

(viii) 46656 =

2 × 2 × 2 × 2 × 2 × 2 3 × 3 × 3 3 × 3 × 3

= 2

3

× 2

3

× 3

3

× 3

3

3

46656

= 2 × 2 × 3 × 3 = 36

(ix) 175616 =

2 × 2 × 2 × 2 × 2 × 2 2 × 2 × 2 7 × 7 × 7

= 2

3

× 2

3

× 2

3

× 7

3

3

175616

= 2 × 2 × 2 × 7 = 56

(x) 91125 =

3 × 3 × 3 × 3 × 3 × 3 3 × 3 × 3 5 × 5 × 5

= 3

3

× 3

3

× 5

3

3

91125

= 3 × 3 × 5 = 45

Question 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

Solution:

(i) False (ii) True (iii) True (iv) False (v) False (vi) False (vii) True

Question 3

You are told that 1,331 is a perfect cube.

Can you guess without factorization what its cube root is?

Similarly, guess the cube roots of 4913, 12167, 32768.

Answer:

11, 17, 23, 32

Solution:

The given number is 1331.

Step I:

From groups of three digits starting from the rightmost digit of 1331.

The only group of three digits is 331.

The left out group of digits is just 1.

Step II:

Let’s consider 331.

The digit 1 is at its one’s place

We take the one’s digit of the required cube root as 1.

Step III:

Let’s take the other group, which is 1

Cube of 1 is 1 and cube of 2 is 8.

But 1 lies between 1and 8.

The smaller number among 1 and 2 is 1.

The one’s digit of 1 is 1 itself.

Taking 1 as ten’s digit of the cube root of 1331, the required cube root = 11.

Cube root of 4913.

Step I:

Grouping from the right: 4913

Step II:

Let’s take 913.

The digit 3 is at its one’s place; therefore we take the one’s digit of the required

cube root as 7 (7 × 7 × 7 = 343).

Step III:

Let’s take the other group, whichis 4.

Cube of 1 is 1 and cube of 2 is 8.

But 4lies between 1 and 8.

The smaller number among 1 and 2 is 1.

The one’s place of 1 is 1 itself.

Taking 1 as ten’s place of the cube root of 4913,

The cube root of 4913 = 17

Cube root of 12167

Step I:

Grouping from the right: 12167

Step II:

3 × 3 × 3 = 27, i.e., one’s digit of the cube root is 3.

Step III:

Taking the other group, which is 12, 2 × 2 × 2 = 8, 3 × 3 × 3

= 27 and 8 < 12 < 27

The smaller number among 2 and 3 is 2.

The one’s place of 2 is 2 itself.

Taking 2 as ten’s digit of the cube root of 12167.

The required cube root = 23

Cube root of 32768

Step I:

Grouping from the right: 32768

Step II:

2 × 2 × 2 = 8 i.e., one’s digit of the cube root is 2.

Step III:

Taking the other group, which is 32, 3 × 3 × 3

= 27, 4 × 4× 4 = 64 and27 < 32 < 64.

The smallest number among 3 and 4 is 3.

The one’s digit of 3 is 3 itself.

Taking 3 as ten’s digit of the cube root of 32768, the required cube root = 32.