NCERT Solutions for Class 8 Maths Chapter 7 Cubes and Cube Roots

Class 8 maths chapter 6 question answer - Topic 6.2 Cubes

Q(i) Find the one’s digit of the cube of each of the following numbers.

3331

Answer: The detailed solution for the above-mentioned question is as follows,

Since the given number ends with 1, so the one’s digit of the cube of 3331 will be 1.

Q(ii) Find the one’s digit of the cube of each of the following numbers.

8888

Answer: The detailed solution for the above-mentioned question is as follows

Since the given number ends with 8, so the one’s digit of the cube of 8888 will be 2.

Q(iii) Find the one's digit of the cube of each of the following numbers.

149

Answer: The detailed solution for the above-mentioned question is as follows,

Since the given number has 9 at units place, so the one’s digit of the cube of 149 will be 9.

Q(iv) Find the one’s digit of the cube of each of the following numbers.

1005

Answer: The detailed solution for the above-mentioned questions is as follows

Since the given number ends with 5, so one's digit of its cube will also end with 5.

Q(v) Find the one’s digit of the cube of each of the following numbers.

1024

Answer: The solution to the above-mentioned question is as follows,

The given digit ends with 4. So the one’s digit of the cube of 1024 will be 4.

Q(vi) Find the one’s digit of the cube of each of the following numbers.

77

Answer: The detailed solution for the above-mentioned question is as follows,

The given number ends with 7, so its cube will end with 3.

Q(vii) Find the one’s digit of the cube of each of the following numbers.

5022

Answer: The detailed solution for the above-mentioned question is as follows,

Since the given number ends with 2, its cube will end with 8.

Q(viii) Find the one’s digit of the cube of each of the following numbers.

53

Answer: The detailed solution for the above-mentioned question is as follows,

Since the given number has 3 at units place, so, its cube will end with 7.

NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots: Topic 6.2.1
Subtopic Some Interesting Patterns

Q(a) Express the following numbers as the sum of odd numbers using the above pattern?

$6^{3}$

Answer: The detailed solution for the above-mentioned question is as follows,

$6^3 =$ 216 => 31 + 33 + 35 + 37 + 39 + 41

Q(b) Express the following numbers as the sum of odd numbers using the above pattern?

$8^{3}$

Answer: The detailed solution for the above-mentioned question is as follows

$8^3 =$ 512 => 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71

Q(c) Express the following numbers as the sum of odd numbers using the above pattern?

$7^{3}$

Answer: The detailed solution for the above-mentioned question is as follows

73 = 43 + 45 + 47 + 49 + 51 + 53 + 55

Q(i) The detailed solution of the above-written question is as follows,

Using the above pattern

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Find the value of the following

$7^{3}-6^{3}$

Answer: The value of the following question is:

$7^{3}-6^{3} =$ $7^{3}-6^{3} = 1 + 7\times6\times3 = 1 + 126 = 127$

Q(ii) The detailed solution for all the above-written questions is as follows

Using the above pattern

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Find the value of the following.

$12^{3}-11^{3}$

Answer: $12^{3}-11^{3} = 1 + 12\times11\times3 = 1 + 396 = 397$

Q(iii). Consider the following pattern.

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Using the above pattern, find the value of the following.

$20^{3}-19^{3}$

Answer: The detailed solution for the above-written question is mentioned below,

$20^{3}-19^{3} = 1 + 20\times19\times3 = 1 + 1140 = 1141$

Q(iv) Consider the following pattern.

$2^{3}-1^{3}=1+2\times 1\times 3$

$3^{3}-2^{3}=1+3\times2\times 3$

$4^{3}-3^{3}=1+4\times 3\times 3$

Using the above pattern, find the value of the following.

$51^{3}-50^{3}$

Answer: The detailed solution for the above question is mentioned below

$51^{3}-50^{3} = 1 + 51\times50\times3 = 1 + 7650 = 7651$

NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots Topic 6.2.1
Subtopic Cubes and Their Prime Factors

Q. Which of the following are perfect cubes?

1. 400

2. 3375

3. 8000

4. 15625

5. 9000

6. 6859

7. 2025

8. 10648

Answer: We will find it by prime factorization whether they make a pair of three prime numbers or not.

(1) $400 = 2\times2\times2\times2\times5\times5$ . So not a perfect cube.

(2) $3375 = 3\times3\times3\times5\times5\times5$ . So it is a perfect cube.

(3) $8000 = 2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is a perfect cube.

(4) $15625 = 5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(5) $9000 = 2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.

(6) $6859 = 19\times19\times19$ . So it is a perfect cube.

(7) $2025 = 3\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(8) $10648 = 2\times2\times2\times11\times11\times11$ . So it is a perfect cube.

NCERT Solutions for Class 8 Maths Chapter 6 Cubes and Cube Roots
Topic 6.2.2 Smallest Multiple That is a Perfect Cube

Q. Check which of the following are perfect cubes.

(i) 2700

(ii) 16000

(iii) 64000

(iv) 900

(v) 125000

(vi) 36000

(vii) 21600

(viii) 10,000

(ix) 27000000

(x) 1000.

What pattern do you observe in these perfect cubes?

Answer: The detailed solution for the above-written question is as follows

By prime factorization:

(i) $2700 = 2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(ii) $16000 = 2\times2\times2\times2\times2\times2\times2\times5\times5\times5$ . So it is not a perfect cube.

(iii) $64000 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times5\times5\times5= 80\times80\times80$ . So it is a perfect cube.

(iv) $900 = 2\times2\times3\times3\times5\times5$ . So it is not a perfect cube.

(v) $125000 = 2\times2\times2\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(vi) $36000 = 2\times2\times2\times2\times2\times3\times3\times5\times5\times5$ . So it is not a perfect cube.

(vii) $21600 = 2\times2\times2\times2\times2\times3\times3\times3\times5\times5$ . So it is not a perfect cube.

(viii) $10000 = 2\times2\times2\times2\times5\times5\times5\times5$ . So it is not a perfect cube.

(ix) $27000000 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times5\times5\times5\times5\times5\times5$ . So it is a perfect cube.

(x) $1000 = 2\times2\times2\times5\times5\times5$ . So it is a perfect cube.

We observe that the numbers above which are perfect cube have the number of zeros in multiple of 3.

NCERT Class 8 Solution Exercise: 6.1
Page number: 76, Total questions: 4

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

216

Answer: The detailed solution for the above-written question is as follows

By prime factorization of 216 gives:

$216 = 2\times2\times2\times3\times3\times3$

Since prime numbers are present in pairs of three, so the given number is a perfect cube.

Q.1(ii) Which of the following numbers are not perfect cubes?

128

Answer: We have 128. By prime factorization we get,

$128 = 2\times2\times2\times2\times2\times2\times2$

Since the prime numbers are not in pairs of three, so the given number is not a perfect cube.

Q.1(iii) Which of the following numbers are not perfect cubes?

1000

Answer: The detailed solution for the above written is as follows

By prime factorization of 1000 we get :

$1000 = 2\times2\times2\times5\times5\times5$ .

So the given number is a perfect cube.

Q.1(iv) Which of the following numbers are not perfect cubes?

100

Answer: The detailed solution for the above-written question is as follows

By prime factorization of 100 :

$100 = 2\times2\times5\times5$ .

Since prime numbers are not in pair of three so given number is not a perfect cube.

Q.1(v) Which of the following numbers are not perfect cubes?

46656

Answer: We have 46656, by prime factorisation:

$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$ .

Since prime numbers are in group of three. So the given number is a perfect cube.

Q.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

Answer: This can be found by knowing about the prime factors of the number.

(i) 243 : $3\times3\times3\times3\times3$ .

So it must be multiplied by 3.

(ii) 256 : $2\times2\times2\times2\times2\times2\times2\times2$

So the given number must be multiplied by 2 to make it a perfect cube.

(iii) 72 : $2\times2\times2\times3\times3$

So 72 must be multiplied by 3 to make it a perfect cube.

(iv) 675 : $3\times3\times3\times5\times5$

So it should be multiplied by 5.

(v) 100 : $2\times2\times5\times5$

So it should be multiplied by 10.

Q.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

Answer: By prime factorization of given numbers:

(i) 81: $3\times3\times3\times3$

So given number needs to be divided by 3 to get a perfect cube.

(ii) 128: $2\times2\times2\times2\times2\times2\times2$ .

So the given number needs to be divided by 2 to get a perfect cube.

(iii) 135: $3\times3\times3\times5$

So the given number needs to be divided by 5 to get a perfect cube.

(iv) 192: $2\times2\times2\times2\times2\times2\times3$

So the given number needs to be divided by 3 to get a perfect cube.

(v) 704: $2\times2\times2\times2\times2\times2\times11$

So the given number needs to be divided by 11 to get a perfect cube.

Q.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: Volume of cuboid is $5\times2\times5 = 2\times5\times5$ $cm^3$

To make it a cube need to make this a perfect cube number.

So we need $2\times2\times5$ cuboids or 20 cuboids.

Q. State true or false:

for any integer $m,m^{2}< m^{3}.$ Why?

Answer: The detailed solution for the above-written question is as follows.

False.

$m^2 < m^3$

Or, $0 < m^3 - m^2$

Or, $m^3 - m^2>0$

Or $m^2\left ( m-1 \right )>0$

Now, put any number less than 1, we see that this relation doesn't hold.

So for m<1 this condition is not true.

NCERT Class 8 Solution Exercise: 6.2
Page number: 77, Total questions: 2

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

64

Answer: The detailed solution for the above-written question is as follows

Prime factorization of 64 gives :

$64 = 2\times2\times2\times2\times2\times2$

So its cube root is $2\times2$ = 4

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

512

Answer: By prime factorisation of 512:

$512 = 2\times2\times2\times2\times2\times2\times2\times2\times2$

So its cube root is $2\times2\times2 = 8$

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

10648

Answer: The detailed solution for the above-written question is as follows

Prime factorization of 10648 gives :

$10648 = 2\times2\times2\times11\times11\times11$

So, its cube root is 22.

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

27000

Answer: The detailed solution for the above-written question is as follows

By the prime factorization method, we get :

$27000 = 2\times2\times2\times3\times3\times3\times5\times5\times5$

So its cube root is 30.

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

15625

Answer: The detailed solution for the above-written question is as follows

By prime factorization:

$15625 = 5\times5\times5\times5\times5\times5$

So its cube root is 25.

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

13824

Answer: The detailed solution for the above-written question is as follows

By prime factorization:

$13824 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$

So its cube root is 24.

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

110592

Answer: The detailed solution for the above-written question is as follows

By prime factorization:

$110592 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times3\times3\times3$

So its cube root is $2\times2\times2\times2\times3$ = 48.

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

46656

Answer: The detailed solution for the above-written question is as follows

By prime factorization, we get :

$46656 = 2\times2\times2\times2\times2\times2\times3\times3\times3\times3\times3\times3$

So its cube root is $2\times2\times3\times3$ = 36.

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

175616

Answer: The detailed solution for the above-written question is as follows

By prime factorization, we get :

$175616 = 2\times2\times2\times2\times2\times2\times2\times2\times2\times7\times7\times7$

So its cube root is $2\times2\times2\times7$ = 56.

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

91125

Answer: The detailed solution for the above-written question is as follows

By prime factorization, we get :

$91125 = 3\times3\times3\times3\times3\times3\times5\times5\times5$

So its cube root is $3\times3\times5$ = 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 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.

Answer: (i) False. The cube of an odd number can never be even.

(ii) True. A perfect cube number ends with zeros multiples of three.

(iii) False. We can say only about the unit's place.

(iv) False. Cube of numbers which end with 2 end with 8.

(v) False. It can never be.

(vi) False. It can never be. It can be proved by taking examples.

(vii) True. e.g. 1,2