## RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

**Other Exercises**

- RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
- RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
- RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
- RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
- RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

**Question 1.**

**Find the cubes of the following numbers:**

**(i) 7**

**(ii) 12**

**(iii) 16**

**(iv) 21**

**(v) 40**

**(vi) 55**

**(vii) 100**

**(viii) 302**

**(ix) 301**

**Solution:
**(i) (7)

^{3}= 7 x 7 x 7 = 343

(ii) (12)

^{3}= 12 x-12 x 12 = 1728

(iii) (16)

^{3}= 16 x 16 x 16 = 4096

(iv) (21)

^{3}= 21 x 21 x 21 = 441 x 21 =9261

(v) (40)

^{3}= 40 x 40 x 40 = 64000

(vi) (55)

^{3}= 55 x 55 x 55 = 3025 x 55 = 166375

(vii) (100)

^{3}= 100 x 100 x 100 =1000000

(viii)(302)

^{3}= 302 x 302 x 302 = 91204 x 302 = 27543608

(ix) (301)

^{3}= 301 x 301 x 301 = 90601 x 301 =27270901

**Question 2.
**

**Write the cubes of all natural numbers between 1 and 10 and verify the following statements :**

**(i) Cubes of all odd natural numbers are odd.**

**(ii) Cubes of all even natural numbers are even.**

**Solution:**

Cubes of first 10 natural numbers :

(1)

^{3}= 1 x 1 x 1 = 1

(2)

^{3}= 2 x 2 x 2 = 8

(3)

^{3}= 3 x 3 x 3 = 27

(4)

^{3}= 4 x 4 x 4 = 64

(5)

^{3}= 5 x 5 x 5 = 125

(6)

^{3}= 6 x 6 x 6 = 216

(7)

^{3}= 7 x 7 x 7 = 343

(8)

^{3}= 8 x 8 x 8 = 512

(9)

^{3}= 9 x 9 x 9= 729

(10)

^{3}= 10 x 10 x 10= 1000

We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.

**Question 3.
**

**Observe the following pattern :**

Write the next three rows and calculate the value of 1

^{3}+ 2

^{3}+ 3

^{3}+…. + 9

^{3}+ 10

^{3}by the above pattern.

**Solution:**

We see the pattern

**Question 4.
**

**Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings :**

**The cube of a natural number which is a multiple of 3 is a multiple of 27′**

**Solution:**

5 natural numbers which are multiples of 3

3,6,9,12,15.

(3)

^{3}= 3 x 3 x 3 = 27

Which is multiple of 27

(6)

^{3}= 6 x 6 x 6 = 216 ÷ 27 = 8

Which is multiple of 27

(9)

^{3}= 9 x 9 x 9 = 729 + 27 = 27

Which is multiple of 27

(12)

^{3}= 12 x 12 x 12 = 1728 ÷ 27 = 64

Which is multiple of 27

(15)

^{3}= 15 x 15 x 15 = 3375 ÷ 27 = 125

Which is multiple of 27

Hence, cube of multiple of 3 is a multiple of 27

**Question 5.
**

**Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following :**

**‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.**

**Solution:**

3n + 1

Let n = 1, 2, 3, 4, 5, then

If n = 1, then 3n +1= 3 x 1+1= 3+1= 4

If n = 2, then 3n +1=3 x 2+1=6+1=7

If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10

If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13

If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16

Now

(4)

^{3}= 4 x 4 x 4 = 64

Which is \(\frac {64 }{ 3 }\)=21, Remainder = 1

(7)

^{3}= 7 x 7 x 7 = 343

Which is \(\frac {343 }{ 3 }\) =114, Remainder = 1

(10)

^{3}= 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1

(13)

^{3}= 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1

(16)

^{3}= 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1

Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1

**Question 6.
**

**Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following :**

**‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’.**

**Solution:**

Natural numbers of the form 3n + 2, when n

is a natural number i.e. 1, 2, 3, 4, 5,………….

If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5

If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8

If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11

If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14

and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17

Now (5)

^{3}= 5 x 5 x 5 = 125

125 + 3 = 41, Remainder = 2

(8)

^{2}= 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2

(11)

^{3}= 11 x 11 x 11 = 1331

1331 + 3 = 443, Remainder = 2

(14)

^{3}= 14 x 14 x 14 = 2744

2744 + 3 = 914, Remainder = 2

(17)

^{3}= 17 x 17 x 17 = 4913

4913 = 3 = 1637, Remainder = 2

We see the cube of the natural number of the

form 3n + 2 is also a natural number of the

form 3n + 2.

**Question 7.
**

**Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following :**

**‘The cube of a multiple of 7 is a multiple of 7**

^{3′}.^{ }

**Solution:**

5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35

(7)

^{3}= (7)

^{3}which is multiple of 7

^{3 }(14)

^{3}= (2 x 7)3 = 2

^{3}x 7

^{3}, which is multiple of 7

^{3 }(21)

^{3}= (3 x 7)3 = 33 x 7

^{3}

_{,}which is multiple of 7

^{3 }(28)

^{3}= (4 x 7)

^{3}= 4

^{3}x 7

^{3}, which is multiple of 7

^{3}(35)

^{3}= (5 x 7)

^{3}= 5

^{3}x 73 which is multiple of 7

^{3 }Hence proved.

**Question 8.
**

**Which of the following are perfect cubes?**

**(i) 64**

**(ii) 216**

**(iii) 243**

**(iv) 1000**

**(v) 1728**

**(vi) 3087**

**(vii) 4608**

**(viii) 106480**

**(ix) 166375**

**(x) 456533**

**Solution:**

**(i) 64 = 2 x 2 x 2 x 2 x 2 x 2**

Grouping the factors in triplets of equal factors, we see that no factor is left

∴ 64 is a perfect cube

**(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3**

Grouping the factors in triplets of equal factors, we see that no factor is left

216 is a perfect cube.

**(iii) 243 = 3 x 3 x 3 x 3 x 3**

Grouping the factors in triplets, we see that two factors 3 x 3 are left

∴ 243 is not a perfect cube.

**(iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5**

Grouping the factors in triplets of equal factors, we see that no factor is left

∴ 1000 is a perfect cube.

**(v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3**

Grouping the factors in triplets of the equal factors, we see that no factor is left

∴ 1728 is a perfect cube,

**(vi) 3087 = 3 x 3 x 7 x 7 x 7**

Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left

∴ 3087 is not a perfect cube.

**(vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3**

Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left

∴ 4609 is not a perfect cube.

**(viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11**

Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left

∴ 106480 is not a perfect cube.

**(ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11**

Grouping the factors in triplets of equal factors, we see that no factor is left

∴ 166375 is a perfect cube.

**(x) 456533 = 7 x 7 x 7 x 11 x 11 x 11**

Grouping the factors in triplets of equal factors, we see that no factor is left

∴ 456533 is a perfect cube.

**Question 9.
**

**Which of the following are cubes of even natural numbers ?**

**216, 512, 729,1000, 3375, 13824**

**Solution:**

We know that the cube of an even natural number is also an even natural number

∴ 216, 512, 1000, 13824 are even natural numbers.

∴ These can be the cubes of even natural number.

**Question 10.
**

**Which of the following are cubes of odd natural numbers ?**

**125, 343, 1728, 4096, 32768, 6859**

**Solution:**

We know that the cube of an odd natural number is also an odd natural number,

∴ 125, 343, 6859 are the odd natural numbers

∴ These can be the cubes of odd natural numbers.

**Question 11.
**

**What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ?**

**(i) 675**

**(ii) 1323**

**(iii) 2560**

**(iv) 7803**

**(v) 107311**

**(vi) 35721**

**Solution:**

**(i) 675 = 3 x 3 x 3 x 5 X 5**

Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet

So, by multiplying by 5, the triplet will be completed.

∴ Least number to be multiplied = 5

**(ii) 1323 = 3 x 3 x 3 x 7 x 7**

Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left

So, multiplying by 7, we get a triplet

∴ The least number to be multiplied = 7

**(iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5**

Grouping the factors in triplet of equal factors, 5 is left.

∴ To complete a triplet 5 x 5 is to multiplied

∴ Least number to be multiplied = 5 x 5 = 25

**(iv) 7803 = 3 x 3 x 3 x 17 x 17**

Grouping the factors in triplet of equal factors, we find the 17 x 17 are left

So, to complete the triplet, we have to multiply by 17

∴ Least number to be multiplied = 17

**(v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11**

Grouping the factors in triplet of equal factors, factor 3 is left

So, to complete the triplet 3 x 3 is to be multiplied

∴ Least number to be multiplied = 3 x 3 = 9

**(vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7**

Grouping the factors in triplet of equal factors, we find that 7 x 7 is left

So, in order to complete the triplets, we have to multiplied by 7

∴ Least number to be multiplied = 7

**Question 12.
**

**By which smallest number must the following numbers be divided so that the quotient is a perfect cube ?**

**(i) 675**

**(ii) 8640**

**(iii) 1600**

**(iv) 8788**

**(v) 7803**

**(vi) 107811**

**(vii) 35721**

**(viii) 243000**

**Solution:**

**(i) 675 = 3 x 3 x 3 x 5 x 5**

Grouping the factors in triplet of equal factors, 5 x 5 is left

5 x 5 is to be divided so that the quotient will be a perfect cube.

∴ The least number to be divided = 5 x 5 = 25

**(ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3**

Grouping the factors in triplets of equal factors, 5 is left

∴ In order to get a perfect cube, 5 is to divided

∴ Least number to be divided = 5

**(iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5**

Grouping the factors in triplets of equal factors, we find that 5 x 5 is left

∴ In order to get a perfect cube 5 x 5 = 25 is to be divided.

∴ Least number to be divide = 25

**(iv) 8788 = 2 x 2 x 13 x 13 x 13**

Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left

∴ In order to get a perfect cube, 2 x 2 is to be divided

∴ Least number to be divided = 4

**(v) 7803 = 3 x 3 x 3 x 17 x 17**

Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left.

So, in order to get a perfect cube, 17 x 17 is be divided

∴ Least number to be divided = 17 x 17 = 289

**(vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11**

Grouping the factors in triplets of equal factors, 3 is left

∴ In order to get a perfect cube, 3 is to be divided

∴ Least number to be divided = 3

**(vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7**

Grouping the factors in triplets of equal factors, we see that 7 x 7 is left

So, in order to get a perfect cube, 7 x 7 = 49 is to be divided

∴ Least number to be divided = 49

**(viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5**

Grouping the factors in triplets of equal factors, 3 x 3 is left

∴ By dividing 3 x 3, we get a perfect cube

∴ Least number to be divided = 3 x 3=9

**Question 13.
**

**Prove that if a number is trebled then its cube is 27 times the cube of the given number.**

**Solution:**

Let x be the number, then trebled number of x = 3x

Cubing, we get:

(3x)

^{3}= (3)

^{3}x

^{3}= 27x

^{3 }27x

^{3}is 27 times the cube of x i.e., of x

^{3}

**Question 14.
**

**What happenes to the cube of a number if the number is multiplied by**

**(i) 3 ?**

**(ii) 4 ?**

**(iii) 5 ?**

**Solution:**

number (x)

^{3}= x

^{3 }(i) If x is multiplied by 3, then the cube of

∴ (3x)

^{3}= (3)

^{3}x x

^{3}= 27x

^{3 }∴ The cube of the resulting number is 27 times of cube of the given number

(ii) If x is multiplied by 4, then the cube of

(4x)

^{3 }= (4)

^{3}x x

^{3}= 64x

^{3 }∴ The cube of the resulting number is 64 times of the cube of the given number

(ii) If x is multiplied by 5, then the cube of

(5x)

^{3}= (5)

^{3}x x

^{3}= 125x

^{3 }∴ The cube of the resulting number is 125 times of the cube of the given number

**Question 15.
**

**Find the volume of a cube, one face of which has an area of 64 m**

^{2}.**Solution:**

Area of one face of a cube = 64 m

^{2 }∴ Side (edge) of cube = √64

= √64 = 8 m

∴ Volume of the cube = (side)

^{3}= (8 m)

^{3 }= 512 m

^{3}

**Question 16.
**

**Find the volume of a cube whose surface area is 384 m**

^{2}.**Solution:**

Surface area of a cube = 384 m

^{2 }Let side = a

Then 6a

^{2}= 384 ⇒ a

^{2}= \(\frac {384 }{ 6 }\)= 64 = (8)

^{2 }∴ a = 8 m

Now volume = a

^{3}= (8)

^{3}m

^{3}= 512 m

^{3}

**Question 17.
**

**Evaluate the following :**

**Solution:**

**Question 18.**

**Write the units digit of the cube of each of the following numbers :**

**31,109,388,833,4276,5922,77774,44447, 125125125.**

**Solution:**

We know that if unit digit of a number n is

= 1, then units digit of its cube = 1

= 2, then units digit of its cube = 8

= 3, then units digit of its cube = 7

= 4, then units digit of its cube = 4

= 5, then units digit of its cube = 5

= 6, then units digit of its cube = 6

= 7, then the units digit of its cube = 3

= 8, then units digit of its cube = 2

= 9, then units digit of its cube = 9

= 0, then units digit of its cube = 0

Now units digit of the cube of 31 = 1

Units digit of the cube of 109 = 9

Units digits of the cube of 388 = 2

Units digits of the cube of 833 = 7

Units digits of the cube of 4276 = 6

Units digit of the cube of 5922 = 8

Units digit of the cube of 77774 = 4

Units digit of tl. cube of 44447 = 3

Units digit of the cube of 125125125 = 5

**Question 19.
**

**Find the cubes of the following numbers by column method :**

**(i) 35**

**(ii) 56**

**(iii) 72**

**Solution:**

**Question 20.
**

**Which of the following numbers are not perfect cubes ?**

**(i) 64**

**(ii) 216**

**(iii) 243**

**(iv) 1728**

**Solution:**

**(i) 64 = 2 x 2 x 2 X 2 x 2 x 2**

Grouping the factors in triplets, of equal factors, we see that no factor is left

∴ 64 is a perfect cube.

**(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3**

Grouping the factors in triplets, of equal factors, we see that no factor is left

∴ 216 is a perfect cube.

**(iii) 243 = 3 x 3 x 3 x 3 x 3**

Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left

∴ 243 is not a perfect cube.

**(iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3**

Grouping the factors m triplets, of equal factors, we see that no factor is left.

∴ 1728 is a perfect cube.

**Question 21.
**

**For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be**

**(a) multiplied so that the product is a perfect cube.**

**(b) divided so that the quotient is a perfect cube.**

**Solution:**

In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left.

**(a)**In order to make it a perfect cube, 3 is to be multiplied which makes a triplet.

**(b)**In order to make it a perfect cube, 3 x 3 or 9 is to be divided.

**Question 22.
**

**By taking three different values of n verify the truth of the following statements :**

**(i) If n is even, then n**

^{3}is also even.**(ii) If n is odd, then n**

^{3}is also odd.**(iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3.**

**(iv) If a natural number n is of the form 3p + 2 then n**

^{3}also a number of the same type.**Solution:**

**(i) n is even number.**

Let n = 2, 4, 6 then

**(a)**n

^{3}= (2)

^{3}= 2 x 2 x 2 = 8, which is an even number.

**(b)**(n)

^{3}= (4)

^{3}= 4 x 4 x 4 = 64, which is an even number.

**(c)**(n)

^{3}= (6)

^{3}= 6 x 6 x 6 = 216, which is an even number.

**(ii) n is odd number.**

Letx = 3, 5, 7

**(a)** (n)^{3} = (3)^{3} = 3 x 3 x 3 = 27, which is an odd number.

**(b)** (n)^{3} = (5)^{3} = 5 x 5 x 5 = 125, which is an odd number.

**(c)** (n)^{3} = (7)^{3} = 7 x 7 x 7 = 343, which is an odd number.

**(iii) If n leaves remainder 1 when divided by 3, then n ^{3} is also leaves 1 as remainder,**

Let n = 4, 7, 10 If n = 4,

then «

^{3}= (4)

^{3}= 4 x 4 x 4 = 64

= 64 + 3 = 21, remainder = 1

If n = 7, then

n

^{3}= (7)

^{3}= 7 x 7 x 7 = 343

343 + 3 = 114, remainder = 1

If n – 10, then

(n)

^{3}= (10)

^{3}= 10 x 10 x 10 = 1000

1000 + 3 = 333, remainder = 1

**(iv) If the natural number is of the form 3p + 2, then n ^{3} is also of the same type**

Let p =’1, 2, 3, then

**(a) If p = 1, then**

n = 3p + 2 = 3 x 1+2=3+2=5

∴ n

^{3}= (5)

^{3}= 5 x 5 x 5 = 125

125 = 3 x 41 + 2 = 3p +2

**(b) If p = 2, then**

n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8

∴ n^{3} = (8)^{3} = 8 x 8 x 8 = 512

∴ 512 = 3 x 170 + 2 = 3p + 2

**(c) If p = 3, then**

n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11

∴ n^{3} = (11)^{3} = 11 x 11 x 11 = 1331

and 1331 =3 x 443 + 2 = 3p + 2

Hence proved.

**Question 23.
**

**Write true (T) or false (F) for the following statements :**

**(i) 0 392 is a perfect cube.**

**(ii) 8640 is not a perfect cube.**

**(iii) No cube can end with exactly two zeros.**

**(iv) There is no perfect cube which ends in 4.**

**(v) For an integer a, a**

^{3}is always greater than a^{2 }> b^{2}**(vi) If a and b are integers such that a**

^{2 }> b^{2}, then a^{3}> b^{3}.**(x) If a**

^{2}ends in an even number of zeros, then a ends in an odd number of zeros.**Solution:**

∵ 5 is left.

(iii)True : A number ending three zeros can be a perfect cube.

(iv) False : v (4)

^{3}= 4 x 4 x 4 = 64, which ends with 4.

(v) False : If n is a proper fraction, it is not possible.

(vi) False : It is not true if a and b are proper fraction.

(vii) True.

(viii) False : as a

^{2}ends in 9, then a

^{3}does not necessarily ends in 7. It ends in 3 also.

(ix) False : it is not necessarily that a

^{3}ends in 25 it can end also in 75.

(x) False : If a

^{2}ends with even zeros, then a

^{3 }will ends with odd zeros but of multiple of 3.

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 are helpful to complete your math homework.

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