CBSE Class 8 Maths Worksheet Chapter 7 Cube and Cube Roots - PDF

1. What is the volume of a cube whose each side is $4 \mathrm{~cm}$ ?

1. 48 cubic $\mathrm{cm}$.

2. 24 cubic $\mathrm{cm}$.

3. 125 cubic cm.

4. 64 cubic cm.

2. What is the value of $\sqrt[3]{512}$

1. 8

2. 6

3. 7

4. 9

3. The value of $\sqrt[3]{343} \times \sqrt[3]{64}$ is

1. 28

2. -28

3. 18

4. -18

4. Which of the following is a perfect cube?

1. 294

2. 496

3. 216

4. 141

5. Find the cube root of 64 by the prime factorisation method.

1. 4

2. 2

3. 6

4. 8

6. What is the value of $\sqrt[3]{\dfrac{-729}{512}}$

1. $\dfrac{-9}{8}$

2. $\dfrac{9}{8}$

3. $\dfrac{-9}{7}$

4. $\dfrac{9}{7}$

7. Fill in the Blanks.

1. $\sqrt[3]{1728}=4 \times$ _____

2. $\sqrt[3]{8 \times \ldots \ldots \ldots  \ldots}=8$

3. The cube root of the number $\mathrm{n}$ is denoted by __________

8. Match the column:

9. Find the cube of $\dfrac{2}{7}$

10. If a cube has a volume of 343 cubic meters, calculate the length of each side.

11. If one side of a cube is $1.21$ meters in length, find its volume.

12. Evaluate: $\sqrt[3]{216}+\sqrt[3]{343}-\sqrt[3]{1321}$

13. Three numbers in the ratio $4: 3: 2$. The sum of their cubes is 334125. Find the numbers.

14. Find the cube of the rational number $4.01$.

15. 6859 is the perfect cube or not?

16. Find the cube root of (-729)

17. Find the volume of a cube whose surface area is 216 square cm.

18. What should we multiply in the “108”number so that it becomes a perfect cube?

19. What should we divide in the “135” number so that it becomes a perfect cube?

20. What minimal multiplier should be used to multiply 3600 so that the result is a perfect cube? Additionally, determine a quotient's cube root.

21. Find the cube root of 512 by the prime factorisation method.

22. Find the smallest number by which 243 must be multiplied to obtain a perfect cube.

23. Find the smallest number by which 81 must be divided to obtain a perfect cube.

24. Find the smallest number by which 128 must be divided to obtain a perfect cube.

25. Parikshit makes a cuboid of plasticine of sides $5 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$. How many such cuboids will he need to form a cube?

Answers to the Worksheet:

1. (d) 64 cubic cm.

Volume of cube = (side)3 = (4)3 = 64cubic cm.

2. (a) 8

Step 1: Find the prime factors of 512

$512=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Step 2: Pair the factors of 512 in a group of three, such that they form cubes.

$512=(2 \times 2 \times 2) \times(2 \times 2 \times 2) \times(2 \times 2 \times 2)$

$512=2^3 \times 2^3 \times 2^3$

$512=8^3$

Step 3: Now, we will apply cube root on both sides to take out the factor (in cubes) as a single term.

$\sqrt[3] {512}=\sqrt[3]{\left(8^3\right)}$

So, here the cube root is eliminated by the cube of 8.

Hence, $\sqrt[3]{512}=8$

3. (a) 28

$\sqrt[3]{343} \times \sqrt[3]{64}=\sqrt[3]{343 \times 64}\left(\because a^m \times b^n=(a \times b)^n\right)$

$=\sqrt[3]{7 \times 7 \times 7 \times 4 \times 4 \times 4}$

$=\left(7^3 \times 4^3\right)^{\dfrac{1}{3}}$

$=\left((7 \times 4)^3\right)^{\dfrac{1}{3}}$    $\because\left(a^m\right)^n=(a)^{m \times n}$

$=(7 \times 4)^{3 \times \dfrac{1}{3}}$

$=28$

4. (c) 216

Factors of 141 = 3 × 47

Factors of 294 = 2 × 7 × 7 × 3

Factors of 216 = 2 × 2 × 2 × 3 × 3 × 3 = $2{}^3 \times 3{}^3$

Factors of 496 = 2 × 2 × 2 × 2 × 31

We see that 216 is a perfect cube. Hence, option c is correct.

5. (a) 4

Prime factorisation of 64 is

$64=2 \times 2 \times 2 \times 2 \times 2 \times 2$

Therefore,

$\sqrt[3]{64}=2 \times 2 =4$

6. (a) $\dfrac{-9}{8}$

7. 

1. $\sqrt[3]{1728} = \sqrt[3]{12^3} = 12 = 4 \times \underline{3}$

2. $\sqrt[3]{8 \times \underline{ 8 \times 8}}$ = 8 

3. The cube root of the number $\mathrm{n}$ is denoted by $ \underline{\sqrt[3]{n}}$

8. Match the column:

Explanation: 

(a) $4^3$ is equal to: $4 \times 4 \times 4$ = 64

(b) The value of $\sqrt[3]{\dfrac{125}{27}}$: 

$\sqrt[3]{\dfrac{125}{27}}$

$=\sqrt[3]{\dfrac{5^3}{3^3}}$

$=\dfrac{5}{3}$

Hence the value of $\sqrt[3]{\dfrac{125}{27}}$ is$=\dfrac{5}{3}$.

(c) The Hardy- Ramanujan Number is the smallest number which can be expressed as the sum of two different cubes in two different ways. And it is  1729

$1729=1728+1=12^3+1^3$

$1729=1000+729=10^3+9^3$

(d) 675 = 3 × 3 × 3 × 5 × 5

By grouping the factors in triplets of equal factors,

675 = (3 × 3 × 3) × 5 × 5

Here, 5 cannot be grouped into triplets of equal factors.

∴ We will multiply 675 by 5 to get perfect cube. 

9. Cube of $\dfrac{2}{7}$

$\Rightarrow \left(\dfrac{2}{7}\right)^3$

$\Rightarrow \dfrac{2 \times 2 \times 2}{7 \times 7 \times 7}$

$\Rightarrow \dfrac{8}{343}$

10. We know that,

Volume $=(\text { Side })^3$

$343=(\text { Side })^3$

Side $=\sqrt[3]{343}$

Side $=\sqrt[3]{7 \times 7 \times 7}$

Side $=7$ meter.

11. We know that,

Volume $=(\text { Side })^3$

Volume $=(1.21)^3$

Volume $=1.21 \times 1.21 \times 1.21$

Volume $=1.7715$ cubic meter

12. $\sqrt[3]{216}+\sqrt[3]{343}-\sqrt[3]{1321}$

$\Rightarrow \sqrt[3]{6 \times 6 \times 6}+\sqrt[3]{7 \times 7 \times 7}-\sqrt[3]{11 \times 11 \times 11}$

$= 6+7-11$

$= 13-11$

$= 2$

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

According to the question,

$\Rightarrow (4 x)^3+(3 x)^3+(2 x)^3=334125$

$\Rightarrow 64 x^3+27 x^3+8 x^3=334125$

$\Rightarrow 99 x^3=334125$

$\Rightarrow x^3=\dfrac{334125}{99}$

$\Rightarrow x^3=3375$

$\Rightarrow x=15$

Numbers are

$\Rightarrow 4 x=4 \times 15=60$

$\Rightarrow 3 x=3  \times 15=45$

$\Rightarrow 2 x=2  \times 15=30$

14. $(4.01)^3$

$=\Rightarrow 4.01 \times 4.01 \times 4.01$

$=\dfrac{401}{100} \times \dfrac{401}{100} \times \dfrac{401}{100}$

$=\dfrac{644812}{1000000}$

$=\Rightarrow 64.4812$

15. Using Prime factorisation method:

So,

$6859=19 \times 19 \times 19$

Then, there are no factors left behind when the prime factors of 6859 are arranged into triples.

Therefore, 6859 is a perfect cube.

16. $\sqrt[3]{-729}$

$\Rightarrow \sqrt[3]{-729}=\sqrt{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times-1}$

$\Rightarrow \sqrt[3]{-729}=\sqrt{-9 \times 9 \times 9}$

$\Rightarrow \sqrt[3]{-729}=-9$

17. Surface area of the cube is 216 square $\mathrm{cm}$.

Let the length of each edge is $x$

The surface area of a cube is $6 x^2$

According to the question,

$\Rightarrow 6 x^2=216$

$\Rightarrow x^2=\dfrac{216}{6}$

$\Rightarrow x^2=36$

$\Rightarrow x=6 \mathrm{~cm}$

We know that,

Volume of cube $=x^3$ cubic $\mathrm{cm}$.

The volume of cube $=6^3$

Volume of cube $=6 \times 6 \times 6$

Volume of cube $=216$ cubic $\mathrm{cm}$.

18. Using Prime Factorisation method:

$108=2 \times 2 \times 3 \times 3 \times 3$

Factoring in this number shows that the pair of 3 is being formed in the given number but the pair of 2 is not being formed, so we have to multiply by 2 .

$108 \times 2=216$

19.  Using Prime Factorisation method:

$135=3 \times 3 \times 3 \times 5$

Factoring in this number shows that the pair of 3 is being formed in the given number, but the pair of 5 is not being formed, so we have to divide by 5.

$135 \div 5=27$

20. $3600=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$

There are no pairs of 2, 3, and 5 in this factorisation, so to make it a triplet, we need to multiply two times 2 and one times 3 and 5 .

$=3600 \times 2 \times 2 \times 3 \times 5$

$=3600 \times 60$

$=216000$

So, $\sqrt[3]{21600} = 60$

This way, we can multiply the given number by 60 to make a perfect cube.

21. Prime factorisation of 512 is

$512=\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}$

$\therefore \sqrt[3]{512}=2 \times 2 \times 2=8$

22. The prime factorisation of 243 is 

$243=3 \times 3 \times 3 \times 3 \times 3$ 

Here, two $3’s$ are extra which are not in a triplet. To make 243 a cube, one more 3 is required.

In that case, $243 \times 3 =3 \times 3 \times 3 \times 3 \times 3 \times 3 = 729$ is a perfect cube.

Therefore, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3 .

23. $81=3 \times 3 \times 3 \times 3$. 

Here, one 3 is extra which is not in a triplet. Dividing 81 by 3 , will make it a perfect cube.

Thus, $81 \div 3=27=3 \times 3 \times 3$ is a perfect cube. 

Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

24. $128=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$. 

Here, one 2 is extra which is not in a triplet. If we divide 128 by 2 , then it will become a perfect cube. 

Thus, $128 \div 2=64= (2 \times 2 \times 2) \times (2 \times 2 \times 2)$ is a perfect cube. 

Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

25. Some cuboids of size $5 \times 2 \times 5$ are given. 

These cuboids, when arranged to form a cube, the side of this cube is so formed that it will be a common multiple of the sides (i.e., 5, 2, and 5) of the given cuboid.

Finding the LCM of 5, 2 and 5, we get 10. Thus, a cube of $10 \mathrm{~cm}$ side needs to be made. 

For this arrangement, we have to put 2 cuboids along with their length, 5 along with its width, and 2 along with their height. 

Therefore, the total cuboids required according to this arrangement $=2 \times 5 \times 2=20$ With the help of 20 cuboids of such measures, the required cube is formed.

Otherwise,

Volume of the cube of sides $5 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}=5 \mathrm{~cm} \times 2 \mathrm{~cm} \times 5 \mathrm{~cm}=(5 \times 5 \times 2) \mathrm{cm}^3$

Here, two 5 s and one 2 are extra which are not in a triplet. If we multiply this expression by $2 \times 2 \times 5=20$, then it will become a perfect cube. 

Thus, $(5 \times 5 \times 2 \times 2 \times 2 \times 5)=(5 \times 5 \times 5 \times 2 \times 2 \times 2)=1000$ is a perfect cube. 

Hence, 20 cuboids of $5 \mathrm{~cm}, 2 \mathrm{~cm}, 5 \mathrm{~cm}$ are required to form a cube.

Benefits of Class 8 Maths Cube and Cube Roots Worksheet

The Class 8 Maths Cube and Cube Roots worksheet is answered simply and engagingly. The problems from Chapter 7 of the CBSE Class 8 textbook have been thoroughly and step-by-step handled by our experienced team, which helps the students solidify their understanding.

The Class 8 Maths Cube and Cube Roots answer is a comprehensive resource for teachers of students in Class 8. Through the use of these Maths worksheets, educators can gain knowledge and understanding about the development of students as well as methods and techniques for teaching them effectively.

Examples of Class 8 Cube and Cube Roots

Here are some common examples of Cube and Cube Roots exercises:

Q. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. 

(a) 243 

Ans: The prime factorization of 243 is

243 = 3 × 3 × 3 × 3 × 3

Here, two 3s are extra, which are not in a triplet. To make 243 a cube, one more 3 is required. 

In that case, 

243 × 3 = 3×3×3×3×3×3 = 729

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

Therefore, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(b) 256 

Ans: The prime factorization of 

256 = 2×2×2×2×2×2×2×2

Here, two 2s are extra, which are not in a triplet. To make 256 a cube, one more is required. Then, we obtain 

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

256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 , which is a perfect cube.

Therefore, 2 is the smallest natural number by which 256 should be multiplied to make it a perfect cube. 

(c) 675

Ans:

675 = 3 × 3 × 3 × 5 × 5

675 = 3 × 3 × 3 × 5 × 5 Here, two 5s are extra, which are not in a triplet. To make 675 a perfect cube, one more 5 is required. 

Then, we obtain 

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375

675 × 5 = 3 × 3 × 3 × 5 × 5 × 5 = 3375 (a perfect cube). 

Therefore, 5 is the smallest natural number by which 675 should be multiplied to make it a perfect cube. 

Q. True or False Type Questions.

(i) A perfect cube never ends with two zeroes. 

Answer: True.

Reason: Perfect cube ends with a fixed number of zeroes, always a perfect multiple of 3.

e.g., the cube of 10 is 1000, having three zeroes at the end. The cube of 100 is 1000000, having six zeroes at the end. 

(ii) If a square of any number ends with 5, its cube will always end with 25.

Answer: False. 

Reason: It is not the necessary condition that if any square of a number ends with 5, then its cube will certainly end with 25. 

e.g., the square of 25 =625

625 has its unit digit as 5. 

The cube of 25 is 15,625. 

However, the square of 35 is 1,225 and has its unit place digit as five, but the cube of 35 is 42875, which does not end with 25.

Q. We know that 1,331 is the ideal cube. Can you figure out what its cube root is without factorization? Try to determine the cube roots of 4913, 12167, and 32768.

Solution:

(i) When we combine the digits, we obtain 1 and 331.

Assuming that the unit digit of the cube is 1, we can infer that the unit digit of the cube root is also 1.

We obtain 1 as the cube root of 1331's unit digit.

The number in the second group matches the cube of 1 exactly.

Our cube root's tenth digit is used as the smallest number's unit place.

As we know that the unit digit of a number's cube whose unit location is the number one is 1

∴ \[\sqrt[3]{1331 } \] = 11

(ii) The result of grouping the digits is 4 and 913.

Since the unit digit for the cube is 3, we can infer that the unit digit for the cube root is 7. This gives us 7 as the unit digit for the cube root of 4913. As 1 > 4 > 8, we know that 13 = 1 and 23 = 8.

So, we assume 1 to be the tenth digit of the cube root.

∴ \[ \sqrt[3]{4913} \] = 17

(iii) The results of grouping the numerals are 12 and 167.

As the unit digit of the cube is 7, we know that the unit digit of the cube root is 3.

The unit digit of 12167's cube root is 3. Since 8 > 12 > 27, we know that 23 = 8 and 33 = 27.

So, 2 is used as the tenth digit of the cube root.

∴ \[\sqrt[3]{12167}\] = 23

(iv) By grouping the digits, we will get 32, 768.

As We know, the unit digit of the cube = 8,

 and the unit digit of the cube root = 2.

Therefore, 2 is the unit digit of the cube root of 32768. We know 3³ = 27 and 4³ = 64 , 27 > 32 > 64.

Thus, 3 is taken as ten digits of cube root.

∴ \[\sqrt[3]{32768} \] = 32

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Students can find Vedantu's Cubes And Cube Roots Class 8 PDF about the difference between cubes and cube roots with a fun and easy explanation. Cubes and Cube Roots worksheets for Class 8 with answers also explain to the students how to use the prime factorization method to determine an integer's cubes and cube roots. The process for determining a cube number's cube root is also covered in this chapter with cubes and cubes roots for integers with no more than three digits after completing the chapter "Cubes and Cube Roots."