### NCERT Solutions for Class 8 Maths Chapter 7 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

**Solution:**

(i) **The Prime factor of 216 is:**

216 = 2 × 2 × 2 × 3 × 3 × 3

In the above factor , 2 and 3 have fashioned a team of three.

Thus, 216 is a perfect cube.

(ii) **The Prime factor of 128 is:**

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Here, 2 is left without making a group of three.

Thus 128 is not a perfect cube.

(iii) **The Prime factor of 1000 is:**

1000 = 2 × 2 × 2 × 5 × 5 × 5

Here, no number is left for making a group of three.

Thus, 1000 is a perfect cube.

(iv) **The Prime factor of 100 is:**

100 = 2 × 2 × 5 × 5

Here 2 and 5 have not formed a group of three.

Thus, 100 is not a perfect cube.

(v) **The Prime factor of 46656 is:**

46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Here 2 and 3 have formed the groups of three.

Thus, 46656 is a perfect cube.

**Question 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)** The Prime factor of 243 is:**

243 = 3 × 3 × 3 × 3 × 3 = 3^{3} × 3 × 3

Here, no 3 is required to make 3 × 3 a group of three, i.e., 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(ii) **The Prime factor of 256 is:**

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^{3} × 2^{3} × 2 × 2

Here, a number 2 is needed to make 2 × 2 a group of three, i.e., 2 × 2 × 2

Thus, the required smallest number to be multiplied is 2.

(iii) **The Prime factor of 72 is:**

72 = 2 × 2 × 2 × 3 × 3 = 2^{3} × 3 × 3

Here, a number 3 is required to make 3 × 3 a group of three, i.e. 3 × 3 × 3

Thus, the required smallest number to be multiplied is 3.

(iv) **The Prime factor of 675 is:**

675 = 3 × 3 × 3 × 5 × 5 = 3^{3} × 5 × 5

Here, a number 5 is required to make 5 × 5 a group of three to make it a perfect cube, i.e. 5 × 5 × 5

Thus, the required smallest number is 5.

(v) **The Prime factor of 100 is:**

100 = 2 × 2 × 5 × 5

Here, number 2 and 5 are needed to multiplied 2 × 2 × 5 × 5 to make it a perfect cube, i.e., 2 × 2 × 2 × 5 × 5 × 5

Thus, the required smallest number to be multiplied is 2 × 5 = 10.

**Question 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) 92

(v) 704

**Solution:**

(i) **The Prime factor of 81 is:**

81 = 3 × 3 × 3 × 3 = 3^{3} × 3

Here, a number 3 is the number by which 81 is divided to make it a perfect cube,

i.e., 81 ÷ 3 = 27 which is a perfect cube.

Thus, the required smallest number to be divided is 3.

(ii)** The Prime factor of 128 is:**

128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^{3} × 2^{3} × 2

Here, a number 2 is the smallest number by which 128 is divided to make it a perfect cube,

i.e., 128 ÷ 2 = 64 which is a perfect cube.

Thus, 2 is the required smallest number.

(iii) **The Prime factor of 135 is:**

135 = 3 × 3 × 3 × 5 = 3^{3} × 5

Here, 5 is the smallest number by which 135 is divided to make a perfect cube,

i.e., 135 ÷ 5 = 27 which is a perfect cube.

Thus, 5 is the required smallest number.

(iv) **The Prime factor of 192 is:**

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2^{3} × 2^{3} × 3

Here, Number 3 is the smallest no by which 192 is divided to make it a perfect cube,

i.e., 192 ÷ 3 = 64

which is a perfect cube.

Thus, 3 is the required smallest number.

(v)** The Prime factor of 704 is:**

704 = 2 × 2 × 2 × 2 × 2 × 2 × 11 = 2^{3} × 2^{3} × 11

Here, Number 11 is the small number by which 704 is divided to make it a perfect cube,

i.e., 704 ÷ 11 = 64

which is a perfect cube.

Thus, 11 is the required smallest number.

**Question 4.** Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will be needed to form a cube?

**Solution:** The sides of the cuboid are given as 5 cm, 2 cm and 5 cm.

Volume of cuboid = 5 cm × 2 cm × 5 cm = 50 cm^{3}

For the prime factor of 50,

we have 50 = 2 × 5 × 5

To make it a perfect cube, we must have

2 × 2 × 2 × 5 × 5 × 5

= 20 × (2 × 5 × 5)

= 20 × volume of the given cuboid (5 cm, 2 cm and 5 cm)

Thus, Required number of cuboids = 20.

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

**Solution:**

**Question 2. **State True or False.

- Cube of an odd number is even.
- A perfect cube does not end with two zeros.
- If the square of a number ends with 5, then its cube ends with 25.
- There is no perfect cube which ends with 8.
- The cube of a two digit number may be a three digit number.
- The cube of a two digit number may have seven or more digits.
- The cube of a single digit number may be a single digit number.

**Solution:**

- False – Cube of any odd number is always odd, e.g., (7)
^{3}= 343 - True – A perfect cube does not end with two zeros.
- True – If a square of a number ends with 5, then its cube ends with 25, e.g., (5)
^{2}= 25 and (5)^{3}= 625 - False – (12)
^{3}= 1728 (ends with 8) - False – (10)
^{3}= 1000 (4-digit number) - False – (99)
^{3}= 970299 (6-digit number) - True – (2)
^{3}= 8 (1-digit number)

**Question 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:** By grouping the digits, we get 1 and 331

The given perfect cube = 1 331

Forming groups of three from the rightmost digits of 1 331

IInd group = 1

1st group = 331

One’s digit in first group = 1

One’s digit in the required cube root may be 1.

The second group has only 1.

Estimated cube root of 1331 = 11

Thus 1331−−−−√3 = 11

**(i) Given perfect cube = 4 913**

Forming groups of three from the right most digit of 4 913

IInd group = 4

1st group = 913

One’s place digit in 913 is 3.

One’s place digit in the cube root of the given number may be 7.

Now in IInd group digit is 4

1^{3} < 4 < 2^{3}

Ten’s place must be the smallest number 1.

Thus, the estimated cube root of 4913 = 17.

**(ii) Given perfect cube = 12 167**

Forming group of three from the rightmost digits of 12 167

We have IInd group = 12

1st group = 167

The ones place digit in 167 is 7.

One’s place digit in the cube root of the given number may be 3.

Now in Ilnd group, we have 12

2^{3} < 12 < 3^{3}

Ten’s place of the required cube root of the given number = 2.

Thus, the estimated cube root of 12167 = 23.

**(iii) Given perfect cube = 32 768**

Forming groups of three from the rightmost digits of 32 768, we have

IInd group = 32

1st group = 768

One’s place digit in 768 is 8.

One’s place digit in the cube root of the given number may be 2.

Now in IInd group, we have 32

3^{3} < 32 < 4^{3}

Ten’s place of the cube root of the given number = 3.

Thus, the estimated cube root of 32768 = 32.

**NCERT Solutions for Class 8 Maths Chapter 7- Cubes and Cube roots Summary**

The NCERT solutions for Class 8 Maths Chapter 7 Cubes and Cube roots is provided by knowledge Glow which contains the answers for all the questions present in the chapter. The Chapter 7 Cubes and Cube roots contains 2 exercises, in which, Ex 7.1 deals with Cubes and Ex 7.2 deals with Cube roots. Here you will see of what the chapter discusses on “**Cubes and Cube Roots**”.

Numbers like 1729, 4104, 13832, are known as Hardy – Ramanujan Numbers. They can be expressed as sum of two cubes in two different ways.

Numbers obtained when a number is multiplied by itself three times are known as cube numbers.

If in the prime factorisation of a number each factor appears three times, then the number is a perfect cube.

All Main topics covered in this chapter include: Exercise 7.1 Introduction & Exercise 7.2 Cubes 7.2.1 Some Interesting Patterns 7.2.2 Smallest multiple that is a perfect cube 7.3 Cube Roots 7.3.1 Cube root through prime factorisation method 7.3.2 Cube root of cube number.

**Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 7**

**Question 1: **What kind of questions are there in NCERT Solutions for Class 8 Maths Chapter 7?

Chapter 7 of NCERT Solutions for Class 8 Maths has a couple of desire questions, descriptive kind of questions, lengthy reply kind questions, quick reply kind questions, fill in the blanks and each day existence examples. By the stop of this chapter college students can extend their trouble fixing abilities and time administration skills. This helps in deciding to buy excessive marks in their finals.

**Question 2:** Is NCERT Solutions for Class 8 Maths Chapter 7 adequate to attend all the questions that come in the board exam?

Yes, it is adequate to clear up all the questions that come in the board examination of NCERT Solutions for Class 8 Maths Chapter 7. Practising this chapter can make them analyze the standards flawlessly. These questions have been devised, as per the NCERT syllabus and the guidelines. This makes the college students to rating appropriate marks in the finals.

**Question 3 : **Is it essential to examine all the matters supplied in NCERT Solutions for Class 8 Maths Chapter 7?

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**Question 4:** If x 2, then (x + 1) – x ?

(a) -1

(b) 0

(c) 1

(d) 2

(e) 3

**Question 5:** What is the value of (x^y) + (y^z)?

(a) (x^y) * y^z

(b) x^y * z^y

(c) x^y * y^z

**Question 6:** Find the area of the region bounded by the graph of f(x) x^2 – 4x + 5 and the line y0?

(a) 25

(b) 20

(c) 15