Class 8 Maths - Cubes and Cube Roots NCERT Solutions

Chapter 7: Cubes and Cube Roots (NCERT Solutions)

Exercise 7.1

Q1. Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656

(i) 216
Prime factorisation: 216 = 2 × 2 × 2 × 3 × 3 × 3
Since all factors appear in triplets (groups of three), it is a perfect cube.

(ii) 128
Prime factorisation: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Grouping gives (2 × 2 × 2) × (2 × 2 × 2) × 2. One factor 2 is left ungrouped, so it is not a perfect cube.

(iii) 1000
Prime factorisation: 1000 = 2 × 2 × 2 × 5 × 5 × 5
All factors appear in triplets, so it is a perfect cube.

(iv) 100
Prime factorisation: 100 = 2 × 2 × 5 × 5
Both 2 and 5 do not form triplets. So it is not a perfect cube.

(v) 46656
Prime factorisation: 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
All factors are beautifully grouped in triplets. It is a perfect cube.

Q2. 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

(i) 243
243 = 3 × 3 × 3 × 3 × 3.
Grouping into triplets: (3 × 3 × 3) × 3 × 3.
To complete the triplet, we need one more 3. Multiply by 3.

(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2.
Grouping: (2 × 2 × 2) × (2 × 2 × 2) × 2 × 2.
To complete the triplet, we need one more 2. Multiply by 2.

(iii) 72
72 = 2 × 2 × 2 × 3 × 3.
Grouping: (2 × 2 × 2) × 3 × 3.
To complete the triplet, we need one more 3. Multiply by 3.

(iv) 675
675 = 3 × 3 × 3 × 5 × 5.
Grouping: (3 × 3 × 3) × 5 × 5.
To complete the triplet, we need one more 5. Multiply by 5.

(v) 100
100 = 2 × 2 × 5 × 5.
No prime factor is in a group of three. We need one more 2 and one more 5.
Hence, we multiply by 2 × 5 = 10.

Q3. 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

(i) 81
81 = 3 × 3 × 3 × 3 = (3 × 3 × 3) × 3.
Divide by 3 to leave only the triplet (perfect cube).

(ii) 128
128 = (2 × 2 × 2) × (2 × 2 × 2) × 2.
Divide by 2.

(iii) 135
135 = 3 × 3 × 3 × 5.
Divide by 5.

(iv) 192
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3.
Divide by 3.

(v) 704
704 = 2 × 2 × 2 × 2 × 2 × 2 × 11.
Divide by 11.

Q4. 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?

Volume of 1 cuboid = Length × Breadth × Height
Volume = 5 × 2 × 5 = 50 cm³.
Prime factorisation of 50 = 2 × 5 × 5.
To make it a perfect cube (which the volume of a cube must be), we need to multiply by enough numbers to complete the triplets.
We need two more 2s and one more 5. Therefore, we must multiply by 2 × 2 × 5 = 20.
Hence, he will need 20 such cuboids to form a cube.

Exercise 7.2

Q1. 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

(i) 64
64 = 2 × 2 × 2 × 2 × 2 × 2.
∛64 = 2 × 2 = 4.

(ii) 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2.
∛512 = 2 × 2 × 2 = 8.

(iii) 10648
10648 = 2 × 2 × 2 × 11 × 11 × 11.
∛10648 = 2 × 11 = 22.

(iv) 27000
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5.
∛27000 = 2 × 3 × 5 = 30.

(v) 15625
15625 = 5 × 5 × 5 × 5 × 5 × 5.
∛15625 = 5 × 5 = 25.

(vi) 13824
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3.
∛13824 = 2 × 2 × 2 × 3 = 24.

(vii) 110592
110592 = 2¹² × 3³ = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3).
∛110592 = 2 × 2 × 2 × 2 × 3 = 16 × 3 = 48.

(viii) 46656
46656 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3).
∛46656 = 2 × 2 × 3 × 3 = 4 × 9 = 36.

(ix) 175616
175616 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (7 × 7 × 7).
∛175616 = 2 × 2 × 2 × 7 = 8 × 7 = 56.

(x) 91125
91125 = (3 × 3 × 3) × (3 × 3 × 3) × (5 × 5 × 5).
∛91125 = 3 × 3 × 5 = 9 × 5 = 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 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.

(i) False. The cube of an odd number is always odd. E.g., 3³ = 27.

(ii) True. A perfect cube must end with zeros in multiples of 3. (i.e. 3, 6, 9 zeros, etc.).

(iii) False. E.g., Let number = 15. Square = 225. Cube = 15³ = 3375 (Ends with 75, not 25).

(iv) False. For example, 2³ = 8. Also 12³ = 1728 ends with 8.

(v) False. The smallest two digit number is 10, and 10³ = 1000, which has four digits.

(vi) False. The largest two digit number is 99, and 99³ = 970299, which has six digits. Therefore, it can have at most six digits.

(vii) True. The cubes of 1 and 2 are 1 and 8 respectively, which are single digit numbers.

Q3. You are told that 1331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

For 1331:
Make groups of 3 digits from right to left: 1 331.
First group is 331, which ends in 1, so the unit digit of cube root is 1.
Second group is 1. The largest cube less than or equal to 1 is 1³ = 1. So tens digit is 1.
Cube root = 11.

For 4913:
Groups: 4 913.
First group 913 ends in 3, so unit digit is 7 (since 7³ ends in 3).
Second group 4. Since 1³ = 1 and 2³ = 8, 4 lies between 1 and 8. So tens digit is 1.
Cube root = 17.

For 12167:
Groups: 12 167.
First group 167 ends in 7, so unit digit is 3.
Second group 12. Since 2³ = 8 and 3³ = 27, 12 lies between 8 and 27. Tens digit is 2.
Cube root = 23.

For 32768:
Groups: 32 768.
First group 768 ends in 8, so unit digit is 2.
Second group 32. Since 3³ = 27 and 4³ = 64, 32 lies between 27 and 64. Tens digit is 3.
Cube root = 32.

Class 8 Maths - Cubes and Cube Roots Practice Questions

Chapter 7: Cubes and Cube Roots (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. Evaluate: (1.2)³

(1.2)³ = 1.2 × 1.2 × 1.2
= 1.44 × 1.2 = 1.728.

Q2. Is 392 a perfect cube? If not, find the smallest natural number by which 392 must be multiplied so that the product is a perfect cube.

Prime factorisation: 392 = 2 × 2 × 2 × 7 × 7
The prime factor 7 does not appear in a group of three.
Therefore, 392 is not a perfect cube.
To make it a perfect cube, we need one more 7. So, the smallest number to multiply is 7.

Q3. Find the cube root of -2744.

∛(-2744) = -∛(2744)
2744 = 2 × 2 × 2 × 7 × 7 × 7
∛2744 = 2 × 7 = 14
Therefore, ∛(-2744) = -14.

Q4. What is the smallest number by which 2560 must be divided so that the quotient is a perfect cube?

Prime factorisation: 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 = (2³) × (2³) × (2³) × 5
The prime factor 5 does not form a triplet.
So, we must divide by 5 to get a perfect cube.

Q5. The volume of a cubical box is 13.824 cubic metres. Find the length of each edge of the box.

Volume = (edge)³ = 13.824 m³
edge = ∛(13.824)
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 2³ × 3³
∛13824 = 2 × 2 × 2 × 3 = 24
Since there are 3 decimal places, the cube root has 1 decimal place.
Edge = 2.4 m.

Q6. Find the cube root of 216/2197.

∛(216/2197) = ∛(216) / ∛(2197)
∛216 = 6
∛2197 = 13 (since 13 × 13 × 13 = 2197)
Answer = 6/13.

Q7. Evaluate: ∛(27 × 2744)

∛(a × b) = ∛a × ∛b
∛27 = 3
∛2744 = 14 (from previous derivation)
Answer = 3 × 14 = 42.

Q8. Find the cube root of -1331/-4096.

-1331/-4096 = 1331/4096
∛(1331) = 11
∛(4096) = 16 (since 16³ = 4096)
Answer = 11/16.

Q9. If ∛a = 8, then what is the value of a?

Squaring or cubing both sides:
Cubing both sides: (∛a)³ = 8³
a = 8 × 8 × 8 = 512.

Q10. Three numbers are in the ratio 1 : 2 : 3. The sum of their cubes is 98784. Find the numbers.

Let the numbers be x, 2x, and 3x.
Given: x³ + (2x)³ + (3x)³ = 98784
x³ + 8x³ + 27x³ = 98784
36x³ = 98784
x³ = 98784 / 36 = 2744
x = ∛2744 = 14
The numbers are 14, 2(14), and 3(14).
Numbers: 14, 28, 42.

Q11. Find the cube root of 0.000001.

0.000001 = 1 / 1000000
∛(1 / 1000000) = 1 / ∛(10⁶)
= 1 / 100 = 0.01.

Q12. The volume of a cube is 9261000 m³. Find the side of the cube.

Volume = a³ = 9261000
a = ∛(9261 × 1000)
a = ∛9261 × ∛1000
9261 = 21³, and 1000 = 10³.
a = 21 × 10 = 210 m.

Q13. Estimate the cube root of 857375.

Group the number: 857, 375.
The rightmost group is 375, ending in 5. Thus, the unit digit of the cube root is 5.
The left group is 857.
9³ = 729 and 10³ = 1000.
Since 729 < 857 < 1000, the tens digit is 9.
Estimated cube root = 95.

Q14. Find the value of ∛(392) × ∛(14).

∛(392 × 14) = ∛(5488)
Prime factorisation of 392 = 2³ × 7²
Prime factorisation of 14 = 2 × 7
Product = (2³ × 7²) × (2 × 7) = 2⁴ × 7³.
This is not a perfect cube explicitly but if asked to simplify: it is 7 × 2 × ∛2 = 14∛2.

Q15. Difference of two perfect cubes is 189. If the cube root of the smaller of the two numbers is 3, find the cube root of the larger number.

Cube root of smaller = 3, so smaller number = 3³ = 27.
Let larger number be x. x - 27 = 189.
x = 189 + 27 = 216.
Cube root of 216 is 6.

Q16. Find the smallest number by which 3087 must be multiplied to make it a perfect cube.

3087 = 3 × 1029 = 3 × 3 × 343 = 3 × 3 × 7 × 7 × 7
= 3² × 7³.
To make it a perfect cube, we need to multiply by one more 3.

Q17. Simplify: ∛( -0.008 × 0.125 ).

∛(-0.008) = -0.2
∛(0.125) = 0.5
Product = -0.2 × 0.5 = -0.10.

Q18. Using estimation, find the cube root of 175616.

Groups: 175 and 616.
616 ends in 6, so the unit digit is 6.
For 175: 5³ = 125 and 6³ = 216. Since 125 < 175 < 216, tens digit is 5.
Estimated cube root = 56.

Q19. Evaluate [(∛729) × (∛343)] / ∛27.

∛729 = 9, ∛343 = 7, ∛27 = 3.
Expression = (9 × 7) / 3 = 63 / 3 = 21.

Q20. Prove that if a number is doubled, then its cube becomes eight times the cube of the given number.

Let the given number be x.
Its cube = x³.
When x is doubled, the new number is 2x.
Its cube = (2x)³ = 8x³.
This is exactly 8 times the cube of the original number. (Proved).

Class 8 Maths - Cubes and Cube Roots Summary

Chapter 7: Cubes and Cube Roots (Concepts & Formulas)

1. Key Terms & Definitions

Cube of a Number: The numbers that are generated by multiplying a number by itself three times are called cube numbers or perfect cubes. E.g., 2 × 2 × 2 = 8, so 8 is a perfect cube.
Perfect Cube: A natural number is said to be a perfect cube if it is the cube of some natural number.
Cube Root: Cube root is the inverse operation of finding cube. The symbol ∛ denotes cube root. E.g., ∛27 = 3.

2. Properties of Cubes of Numbers

The cube of an even natural number is always even. E.g., 4³ = 64.
The cube of an odd natural number is always odd. E.g., 3³ = 27.
If a number ends with digits 1, 4, 5, 6, 9, or 0, its cube also ends with the same digit.
If a number ends with 2, its cube ends with 8, and vice versa.
If a number ends with 3, its cube ends with 7, and vice versa.

3. Prime Factorisation Method

To check if a number is a perfect cube or to find its cube root:

Find the prime factors of the given number.
Group the factors in triplets (groups of three) of equal factors.
If no factors are left ungrouped, the number is a perfect cube.
Take one factor from each triplet and multiply them to get the cube root.

Example: Finding ∛216
216 = 2 × 2 × 2 × 3 × 3 × 3
Triplets: (2 × 2 × 2) and (3 × 3 × 3)
∛216 = 2 × 3 = 6

4. Estimating Cube Roots

For large numbers, you can estimate the cube root through grouping:

Form groups of 3 digits starting from the rightmost digit.
First Group (rightmost): Determines the unit digit of the cube root. E.g., if it ends in 7, the cube root ends in 3.
Second Group: Find the closest perfect cubes it lies between. The smaller cube's root gives the tens digit.

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