RBSE Maths Class 8 Chapter 2: Important Questions and Solutions

RBSE Maths Chapter 2 – Cube and Cube root Class 8 Important questions and solutions can be accessed here. The important questions and solutions of Chapter 2, available at BYJU’S, contain detailed step by step explanations. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 8 solutions.

Chapter 2 of the RBSE Class 8 Maths will help the students to solve problems related to cube and cube root of numbers, the cube of even and odd numbers, some patterns related to cube numbers, cube and their prime factors, determining the cube root by prime factor method, cube root of a perfect cube.

RBSE Maths Chapter 2: Exercise 2.1 Textbook Important Questions and Solutions

Question 1: Find out whether the following numbers are perfect cubes.

[i] 512

[ii] 243

[iii] 1000

[iv] 100

[v] 2700

Solution:

[i]
$$\begin{array}{l}512\\ =2\cdot \:256\\ =2\cdot \:2\cdot \:128\\ =2\cdot \:2\cdot \:2\cdot \:64\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:32\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:16\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:8\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:4\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\\ =2^{3}\cdot 2^{3}\cdot 2^{3}\end{array}$$

The factors are in groups of three. Hence, 512 is a perfect cube.

[ii]
$$\begin{array}{l}243\\ =3\cdot \:81\\ =3\cdot \:3\cdot \:27\\ =3\cdot \:3\cdot \:3\cdot \:9\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\\ =3^{3} \cdot 3^{2}\end{array}$$

The factors are not in groups of three. Hence, 243 is not a perfect cube.

[iii]
$$\begin{array}{l}1000\\ =2\cdot \:500\\ =2\cdot \:2\cdot \:250\\ =2\cdot \:2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{3} \cdot 5^{3}\end{array}$$

The factors are in groups of three. Hence, 1000 is a perfect cube.

[iv]
$$\begin{array}{l}100\\ =2 \cdot \:50\\ =2\cdot \:2\cdot \:25\\ =2\cdot \:2\cdot \:5\cdot \:5\\ =2^{2} \cdot 5^{2}\end{array}$$

The factors are not in groups of three. Hence, 100 is not a perfect cube.

[v]
$$\begin{array}{l}2700\\ =2\cdot \:1350\\ =2\cdot \:2\cdot \:675\\ =2\cdot \:2\cdot \:3\cdot \:225\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:75\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:25\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\\ =2^{2} \cdot 3^{3} \cdot 5^{2}\end{array}$$

The factors are not in groups of three. Hence, 2700 is not a perfect cube.

Question 2: Find the smallest number to be multiplied by the following numbers to get the perfect cube.

[i] 108

[ii] 500

[iii] 5400

[iv] 10584

Solution:

[i]
$$\begin{array}{l}108\\ =2\cdot \:54\\ =2\cdot \:2\cdot \:27\\ =2\cdot \:2\cdot \:3\cdot \:9\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\\ =2^{2} \cdot 3^{3}\end{array}$$

The factor 2 is not in groups of three. So, 2 is the number that has to be multiplied with 108 for it to be a perfect cube.

[ii]
$$\begin{array}{l}500\\ =2\cdot \:250\\ =2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{2} \cdot 5^{3}\end{array}$$

The factor 2 is not in groups of three. So, 2 is the number that has to be multiplied with 250 for it to be a perfect cube.

[iii]
$$\begin{array}{l}5400\\ =2\cdot \:2700\\ =2\cdot \:2\cdot \:1350\\ =2\cdot \:2\cdot \:2\cdot \:675\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:225\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:75\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\\ =2^{3} \cdot 3^{3} \cdot 5^{2}\end{array}$$

The factor 5 is not in groups of three. So, 5 is the number that has to be multiplied with 5400 for it to be a perfect cube.

[iv]
$$\begin{array}{l}10584\\ =2\cdot \:5292\\ =2\cdot \:2\cdot \:2646\\ =2\cdot \:2\cdot \:2\cdot \:1323\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:441\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:147\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:49\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:7\cdot \:7\\ =2^{3} \cdot 3^{3} \cdot 7^{2}\end{array}$$

The factor 7 is not in groups of three. So, 7 is the number that has to be multiplied with 10584 for it to be a perfect cube.

Question 3: Find the smallest number to be divided by the following numbers to get the perfect cube.

[i] 108

[ii] 500

[iii] 5400

[iv] 10584

Solution:

[i]
$$\begin{array}{l}108\\ =2\cdot \:54\\ =2\cdot \:2\cdot \:27\\ =2\cdot \:2\cdot \:3\cdot \:9\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\\ =2^{2} \cdot 3^{3}\end{array}$$

The factor 2 is not in groups of three. So, 4 is the number that has to be divided to make it a perfect cube.

[ii]
$$\begin{array}{l}500\\ =2\cdot \:250\\ =2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{2} \cdot 5^{3}\end{array}$$

The factor 2 is not in groups of three. So, 4 is the number that has to be divided to make it a perfect cube.

[iii]
$$\begin{array}{l}5400\\ =2\cdot \:2700\\ =2\cdot \:2\cdot \:1350\\ =2\cdot \:2\cdot \:2\cdot \:675\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:225\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:75\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\\ =2^{3} \cdot 3^{3} \cdot 5^{2}\end{array}$$

The factor 5 is not in groups of three. So, 25 is the number that has to be divided to make it a perfect cube.

[iv]
$$\begin{array}{l}10584\\ =2\cdot \:5292\\ =2\cdot \:2\cdot \:2646\\ =2\cdot \:2\cdot \:2\cdot \:1323\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:441\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:147\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:49\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:7\cdot \:7\\ =2^{3} \cdot 3^{3} \cdot 7^{2}\end{array}$$

The factor 7 is not in groups of three. So, 49 is the number that has to be divided to make it a perfect cube.

Question 4: Rehan works in a soap factory. He is playing with arranging cubic soap by making cubes. If he has to arrange 216 soap, then how many soaps will occur in the first line of the cube.

Solution:

Total number of soaps = 216

The prime factorisation of 216 is as follows:

$$\begin{array}{l}216\\ =2\cdot \:108\\ =2\cdot \:2\cdot \:54\\ =2\cdot \:2\cdot \:2\cdot \:27\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:9\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\\ =2^{3} \cdot 3^{3}\\ =(2 \times 3)^{3}\\ =(6)^{3}\\ \sqrt[3]{216}=6^{3}\end{array}$$

6 soaps will be present in the first row of the cube.

RBSE Maths Chapter 2: Exercise 2.2 Textbook Important Questions and Solutions

Question 1: State whether the following statements are true or false.

(i) Every even number has an even cube number.

(ii) A perfect cube does not end with double zero(00).

(iii) No one perfect cube ends with 8.

(iv) If the square of any number is ending with 5, then its cube ends with 25.

(v) Cube of single-digit is also a single-digit number.

(vi) Cube of double-digit number is of 4 to 6 digits.

Solution:

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

Question 2: Find the cube roots of the following numbers by the method of estimate and prime factors. Verify the answers.

[i] 64

[ii] 343

[iii] 5832

[iv] 74088

[v] 3375

[vi] 10648

[vii] 46656

[viii] 91125

Solution:

[i] 64

The prime factorisation of 64 is given below.

$$\begin{array}{l}64\\ =2\cdot \:32\\ =2\cdot \:2\cdot \:16\\ =2\cdot \:2\cdot \:2\cdot \:8\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:4\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\\ =2^{3} \cdot 2^{3}\\ =(2 \times 2)^{3}\\ \sqrt[3]{64}=2 \times 2=4\end{array}$$

• Group 3-3 digits starting from the unit place, we get 064.
• The first group is 64 and its first digit 4, which can only be obtained by the cube of unit digit 4 (43 = 64). Hence, 4 is the unit digit of the cube root.
• Therefore, the cube root of 64 is 4.
[ii] 343

The prime factorisation of 343 is given below.

$$\begin{array}{l}343\\ =7\cdot \:49\\ =7\cdot \:7\cdot \:7\\ =7^{3}\\ 343=7 \times 7 \times 7\\ 343=7^{3}\\ \sqrt[3]{343}=7\end{array}$$

• Group 3-3 digits starting from the unit place, we get 343.
• The first group is 343 and its first digit 3, which can only be obtained by the cube of unit digit 7 (73 = 343).
• Therefore, the cube root of 343 is 7.
[iii] 5832

The prime factorisation of 5832 is given below.

$$\begin{array}{l}5832\\ =2\cdot \:2916\\ =2\cdot \:2\cdot \:1458\\ =2\cdot \:2\cdot \:2\cdot \:729\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:243\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:81\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:27\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:9\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\\ =2^{3} \cdot 3^{3} \cdot 3^{3}\\ =(2 \times 3 \times 3)^{3}\\ \sqrt[3]{5832}=2 \times 3 \times 3=18\\\end{array}$$

• Group 3-3 digits starting from the unit place, we get 5 832.
• The first group is 832 and its first digit 2, which can only be obtained by the cube of unit digit 8 (83 = 512). Hence, 8 is the unit digit of the cube root.
• The second group is 5 [greater than the cube of 1 and smaller than the cube of 2, hence
$$\begin{array}{l}1^{3}<5<2^{3}\end{array}$$
. The digit in tens place is 1.
• Therefore, the cube root of 5832 is 18.
[iv] 74088

The prime factorisation of 74088 is given below.

$$\begin{array}{l}74088\\ =2\cdot \:37044\\ =2\cdot \:2\cdot \:18522\\ =2\cdot \:2\cdot \:2\cdot \:9261\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3087\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:1029\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:343\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:7\cdot \:49\\ =2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:7\cdot \:7\cdot \:7\\ =2^{3} \cdot 3^{3} \cdot 7^{3}\\ =(2 \times 3 \times 7)^{3}\ \sqrt[3]{74088}=2 \times 3 \times 7=42\end{array}$$

• Group 3-3 digits starting from the unit place, we get 74 088.
• The first group is 088 and its first digit 8, which can only be obtained by the cube of unit digit 2 (23 = 8). Hence, 2 is the unit digit of the cube root.
• The second group is 74 [therefore
$$\begin{array}{l}4^{3}<74<5^{3}\end{array}$$
. The digit in tens place is 4.
• Therefore, the cube root of 74088 is 42.
[v] 3375

The prime factorisation of 3375 is given below.

$$\begin{array}{l}3375\\ =3\cdot \:1125\\ =3\cdot \:3\cdot \:375\\ =3\cdot \:3\cdot \:3\cdot \:125\\ =3\cdot \:3\cdot \:3\cdot \:5\cdot \:25\\ =3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\cdot \:5\\ =3^{3} \cdot 5^{3}\\ =(3 \times 5)^{3}\\ \sqrt[3]{3375}=3 \times 5=15\end{array}$$

• Group 3-3 digits starting from the unit place, we get 3 375.
• The first group is 375 and its first digit 5, which can only be obtained by the cube of unit digit 5 (53 = 125). Hence, 5 is the unit digit of the cube root.
• The second group is 3 [therefore
$$\begin{array}{l}1^{3}<3<2^{3}\end{array}$$
. The digit in tens place is 1.
• Therefore, the cube root of 3375 is 15.
[vi] 10648

The prime factorisation of 10648 is given below.

$$\begin{array}{l}10648\\ =2\cdot \:5324\\ =2\cdot \:2\cdot \:2662\\ =2\cdot \:2\cdot \:2\cdot \:1331\\ =2\cdot \:2\cdot \:2\cdot \:11\cdot \:121\\ =2\cdot \:2\cdot \:2\cdot \:11\cdot \:11\cdot \:11\\ =2^{3} \cdot 11^{3}\\ =(2 \times 11)^{3}\\ \sqrt[3]{10648}=2 \times 11=22\end{array}$$

• Group 3-3 digits starting from the unit place, we get 10 648.
• The first group is 648 and its first digit 8, which can only be obtained by the cube of unit digit 2 (23 = 8). Hence, 2 is the unit digit of the cube root.
• The second group is 10 [therefore
$$\begin{array}{l}2^{3}<10<3^{3}\end{array}$$
. The digit in tens place is 2.
• Therefore, the cube root of 10648 is 22.
[vii] 46656

The prime factorisation of 46656 is given below.

$$\begin{array}{l}46656\\ =2\cdot \:23328\\ =2\cdot \:2\cdot \:11664\\ =2\cdot \:2\cdot \:2\cdot \:5832\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2916\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:1458\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:729\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:3\cdot \:243\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:81\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:27\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:9\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\\ =2^{3} \cdot 2^{3} \cdot 3^{3} \cdot 3^{3}\\ =(2 \times 2 \times 3 \times 3)^{3}\\ \sqrt[3]{46656}=(2 \times 2 \times 3 \times 3)=36\end{array}$$

• Group 3-3 digits starting from the unit place, we get 46 656.
• The first group is 656 and its first digit 6, which can only be obtained by the cube of unit digit 6 (63 = 216). Hence, 6 is the unit digit of the cube root.
• The second group is 46 [therefore
$$\begin{array}{l}3^{3}<46<4^{3}\end{array}$$
. The digit in tens place is 3.
• Therefore, the cube root of 46656 is 36.
[viii] 91125

$$\begin{array}{l}91125\\ =3\cdot \:30375\\ =3\cdot \:3\cdot \:10125\\ =3\cdot \:3\cdot \:3\cdot \:3375\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:1125\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:375\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:125\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:25\\ =3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\cdot \:5\\ =3^{3} \cdot 3^{3} \cdot 5^{3}\\ =(3 \times 3 \times 5)^{3}\\ \sqrt[3]{91125}=(3 \times 3 \times 5)=45\end{array}$$

• Group 3-3 digits starting from the unit place, we get 91 125.
• The first group is 125 and its first digit 5, which can only be obtained by the cube of unit digit 5 (53 = 125). Hence, 5 is the unit digit of the cube root.
• The second group is 91 [therefore
$$\begin{array}{l}4^{3}<91<5^{3}\end{array}$$
. The digit in tens place is 4.
• Therefore, the cube root of 91125 is 45.

RBSE Maths Chapter 2: Additional Important Questions and Solutions

Question 1. Define Hardi- Ramanujan number.

Solution:

The number which can be expressed as the sum of two cubes in two different ways is called Hardi-Ramanujan numbers. For example, 1729, 4104,13832.

Question 2. What is a cube of a number?

Solution:

When a number is multiplied thrice by itself, the obtained result is called a cube of that number.

Question 3. How can 93 be expressed as the sum of consecutive odd numbers?

Solution:

73 + 75 + 77 + 79 + 81 + 83 + 85 + 87 + 89 = 729 = 93

Question 4: What is the cube root of 8000? Solve by prime factorisation method.

Solution:

$$\begin{array}{l}8000\\ =2\cdot \:4000\\ =2\cdot \:2\cdot \:2000\\ =2\cdot \:2\cdot \:2\cdot \:1000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:500\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:250\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{3} \cdot 2^{3} \cdot 5^{3}\\ =(2 \times 2 \times 5)^{3}\\ \sqrt[3]{8000}=(2 \times 2 \times 5)=20\end{array}$$

Question 5: Find the smallest number by which 1188 can be divided to obtain a perfect cube.

Solution:

The prime factorisation of 1188 is given below.

$$\begin{array}{l}1188\\ =2\cdot \:594\\ =2\cdot \:2\cdot \:297\\ =2\cdot \:2\cdot \:3\cdot \:99\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:33\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:11\\ =2^{2} \cdot 3^{3} \cdot 11\end{array}$$

The factor 2 and 11 are not in groups of three. 44 is the smallest number by which 1188 should be divided to obtain a perfect cube.

Question 6: Find the cube root of 9261. Solve by prime factorisation method.

Solution:

$$\begin{array}{l}9261\\ =3\cdot \:3087\\ =3\cdot \:3\cdot \:1029\\ =3\cdot \:3\cdot \:3\cdot \:343\\ =3\cdot \:3\cdot \:3\cdot \:7\cdot \:49\\ =3\cdot \:3\cdot \:3\cdot \:7\cdot \:7\cdot \:7\\ =3^{3} \cdot 7^{3} =(3 \times 7)^{3}\\ \sqrt[3]{9261}=(3 \times 7)=21\end{array}$$

Question 7: A cuboid of the soil with sides 15 cm, 30 cm and 15 cm are made. Find the number of cuboids that are needed to form a cube.

Solution:

The volume of a cuboid =

$$\begin{array}{l}=15 \times 30 \times 15\\ =3 \times 5 \times 2 \times 3 \times 5 \times 3 \times 5\\ =2 \times (3 \times 3 \times 3 \times) [5 \times 5 \times 5]\\ =2 \times [3^{3} \times 5^{3}]\end{array}$$

In the above prime factorization, 2 is not in groups of three. So, 22 = 4 cuboids are needed to form a cube.

Question 8: Rohan has a cuboidal box with sides 5 cm, 3 cm and 5 cm. How many cuboidal boxes will be required for making one cubical box?

Solution:

The volume of a cuboidal box =

$$\begin{array}{l}=5 \times 3 \times 5\\ =[5 \times 5 \times 3]\\ =45\end{array}$$

45 such cuboidal boxes are required to make one cubical box.

Question 9: The ratio of three numbers is 2:3:4 and the sum of their cubes are 33957. Write the greatest number.

Solution:

Let the three numbers be 2x, 3x and 4x.

$$\begin{array}{l}(2x)^{3}+(3x)^{3}+(4x)^{3}=33957\\ 8x^{3}+27x^{3}+64^{3}=33957\\ 99x^{3}=33957\\ x^{3}=\frac{33957}{99}\\ x^{3}=343\\ x^{3}=7^{3}\\ x=7\end{array}$$

The greatest number is 4x = 4 * 7 = 28.

Question 10: If the volume of a cube is 9261000 m3, then find its side.

Solution:

Let the side of a cube be ‘x’ m.

The volume of the cube =

$$\begin{array}{l}a^{3}\end{array}$$

$$\begin{array}{l}a^{3}\end{array}$$
= 9261000

$$\begin{array}{l}a^{3}\end{array}$$
=
$$\begin{array}{l}a=\sqrt[3]{9261000}\\ a=\sqrt[3]{2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 7\cdot 7\cdot 7}\\ a=2\cdot 3\cdot 5\cdot 7=210\end{array}$$

Its side is 210m.

Question 11: We know that 22 = 4 where 4 = 2 x 2 or 4 = 2 + ?. Similarly, 23 = 8 where 8 = 2 x 2 x 2, is this equal to (2 + 2 + 2)?

Solution:

8 = 2 x 2 x 2 = 23

But 2 + 2 + 2 = 6 which is not equal to 2 x 2 x 2.

So, 23 is not equal to 2 + 2 + 2. Hence it is false.

23 = 2 x 2 x 2 is true.

Question 12: Write the unit digits of the cubes given below.

(i) 1331 (ii) 4444 (iii) 159 (iv) 1005

Solution:

(i) In the given number the unit digit is 1.

Unit digit of the cube of the given number will also be 1.

(ii) In the given number unit digit is 4.

Unit digit of the cube of the given number will also be 4.

(iii) In the given number unit digit is 9.

Unit digit of the cube of the given number will also be 9.

(iv) In the given number unit digit is 5.

Unit digit of the cube of the given number will also be 5.

Question 13: Check whether the cube of 46 is odd or even.

Solution:

The number 46 is an even number.

Cube of the given even number will also be an even number.

Hence, the cube of 46 will be an even number.

Question 14: Observe the following pattern of sums of odd numbers.

$$\begin{array}{l}1=1=1^{3}\\ 3+5=8=2^{3}\\ 7+9+11=27=3^{3}\\ 13+15+17+19=64=4^{3}\\ 21+23+25+27+29=125=5^{3}\end{array}$$

How many consecutive odd numbers amount to the sum of

$$\begin{array}{l}10^{3}\end{array}$$
?

Solution:

Cube of 2 consists of 2 odd numbers 2 and 5.

Cube of 3 consists of 3 odd numbers 7, 9 and 11.

Cube of 4 consists of 4 odd numbers 13, 15, 17 and 19.

Cube of 10 consists of 10 consecutive odd numbers.

Question 15: Following the pattern in question 14, write the following numbers in the form of addition of odd numbers.

$$\begin{array}{l}[i] 7^{3}\\ [ii] 8^{3}\end{array}$$

Solution:

$$\begin{array}{l}[i] 7^{3}=43+45+47+49+51+53+55\\ [ii] 8^{3}=57+59+61+63+65+67+69+71\end{array}$$

Question 16: Check whether the following numbers are perfect cubes.

[i] 2700

[ii] 16000

[iii] 64000

[iv] 900

[v] 12500

Solution:

[i]
$$\begin{array}{l}2700\\ =2\cdot \:1350\\ =2\cdot \:2\cdot \:675\\ =2\cdot \:2\cdot \:3\cdot \:225\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:75\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:25\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:3\cdot \:5\cdot \:5\\ =2^{2} \cdot 3^{3} \cdot 5^{2}\end{array}$$

The prime factors 2 and 5 are not in groups of three. Hence, 2700 is not a perfect cube.

[ii]
$$\begin{array}{l}16000\\ =2\cdot \:8000\\ =2\cdot \:2\cdot \:4000\\ =2\cdot \:2\cdot \:2\cdot \:2000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:1000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:500\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:250\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{3} \cdot 2^{3} \cdot 2 \cdot 5^{3}\end{array}$$

The prime factor 2 is not in a group of three. Hence, 16000 is not a perfect cube.

[iii]
$$\begin{array}{l}64000\\ =2\cdot \:32000\\ =2\cdot \:2\cdot \:16000\\ =2\cdot \:2\cdot \:2\cdot \:8000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:4000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:1000\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:500\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:250\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:125\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\\ =2^{3} \cdot 2^{3} \cdot 2^{3} \cdot 5^{3}\end{array}$$

Perfect cubes for each factor is present. Hence, 64000 is a perfect cube.

[iv]
$$\begin{array}{l}900\\ =2\cdot \:450\\ =2\cdot \:2\cdot \:225\\ =2\cdot \:2\cdot \:3\cdot \:75\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:25\\ =2\cdot \:2\cdot \:3\cdot \:3\cdot \:5\cdot \:5\\ =2^{2} \cdot 3^{2} \cdot 5^{2}\end{array}$$

The prime factors 2, 3 and 5 are not in groups of three. Hence, 900 is not a perfect cube.

[v]
$$\begin{array}{l}12500\\ =2\cdot \:6250\\ =2\cdot \:2\cdot \:3125\\ =2\cdot \:2\cdot \:5\cdot \:625\\ =2\cdot \:2\cdot \:5\cdot \:5\cdot \:125\\ =2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\cdot \:25\\ =2\cdot \:2\cdot \:5\cdot \:5\cdot \:5\cdot \:5\cdot \:5\\ =2^{2} \cdot 5^{3} \cdot 5^{2}\end{array}$$

The prime factors 2 and 5 are not in groups of three. Hence, 12500 is not a perfect cube.