Cubes And Cube Roots

(i) 7

Cube of 7 = 7 × 7 × 7 = 343

(ii) 12

Cube of 12 = 12 × 12 × 12 = 1728

(iii) 16

Cube of 16 = 16 × 16 ×16 = 4096

(iv) 21

Scube of 21 = 21 × 21 × 21 = 9261

(v) 40

Cube of 40 = 40 × 40 × 40 = 64000

(vi) 55

Cube of 55 = 55 × 55 × 55 = 166375

(vii) 100

Cube of 100 = 100 × 100 × 100 = 1000000

(viii) 302

To find cube of 302 we make it in form

=

= 27000000 + 8 + 540000 + 3600 = 27362408.

(ix) 301

To find cube of 301 we make it in form

=

= 27000000 + 1 + 270000 + 900 = 27180601.

Cube of natural numbers upto 10 are as follows.

1³ = 1 × 1 × 1 = 1

2³ = 2 × 2 × 2 = 8

3³ = 3 × 3 × 3 = 27

4³ = 4 × 4 × 4 = 64

5³ = 5 × 5 × 5 = 125

6³ = 6 × 6 × 6 = 216

7³ = 7 × 7 × 7 = 343

8³ = 8 × 8 × 8 = 512

9³ = 9 × 9 × 9 = 729

10³ = 10 × 10 × 10 = 1000

From above results we can see that,

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

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

According to given pattern,

=

Here n = 10, so ,

=

=

First 5 natural numbers which are multiple of 3 are = 3 , 6 , 9 , 12 , 15

Now, cube of them are,

= 3³ = 3 × 3 × 3 = 27

= 6³ = 6 × 6 × 6 = 216

= 9³ = 9 × 9 × 9 = 729

= 12³ = 12 × 12 × 12 = 1728

= 15³ = 15 × 15 × 15 = 3375

We find that all the cubes are divisible by 27,

Therefore, “The cube of a natural number which is a multiple of 3 is a multiple of 27’

First 5 natural numbers in the form of ( 3n + 1) are = 4 , 7 , 10 , 13 , 16

Cube of these numbers are,

= 4³ = 4 × 4 × 4 = 64

= 7³ = 7 × 7 × 7 = 343

= 10³ = 10 × 10 × 10 = 1000

= 13³ = 13 × 13 × 13 = 2197

= 16³ = 16 × 16 × 16 = 4096

We find that all these cubes gives remainder 1 when divided by ‘3’

Hence, statement is true.

First 5 natural numbers in form ( 3n + 2 ) are = 5 , 8 , 11 , 14 , 17

Cubes of these numbers are,

= 5³ = 5 × 5 × 5 = 125

= 8³ = 8 × 8 × 8 = 512

= 11³ = 11 × 11 × 11 = 1331

= 14³ = 14 × 14 × 14 = 2744

= 17³ = 17 × 17 × 17 = 4313

We find that all these cubes give remainder 2 when divided by 3..

Hence statement is true.

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

Cube of these numbers are,

= 7³ = 7 × 7 × 7 = 343

= 14³ = 14 × 14 × 14 = 2744

= 21³ = 21× 21× 21 = 9261

= 28³ = 28 × 28 × 28 = 21952

= 35³ = 35 × 35 × 35 = 42875

We find that all these cubes are multiple of 7³(343) as well.

(i) 64

Making factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶ = (2²)³ = 4³

Hence, it’s a perfect cube.

(ii) 216

Factors of 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³ = 6³

Hence, it’s a perfect cube.

(iii) 243

Factors of 243 = 3 × 3 × 3 × 3 × 3 = 3⁵ = 3³ × 3²

Hence, it’s not a perfect cube.

(iv) 1000

Factors of 1000 = 2 × 2 × 2 × 5 × 5 × 5 = 2³ × 5³ = 10³

Hence, it’s a perfect cube.

(v) 1728

Factors of 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2⁶ × 3³ = (4 × 3 )³ = 12³

Hence, it’s a perfect cube.

(vi) 3087

Factors of 3087 = 3 × 3 × 7 × 7 × 7 = 3² × 7³

Hence, it’s not a perfect cube.

(vii) 4608

Factors of 4608 = 2 × 2 × 3 × 113

Hence, it’s not a perfect cube.

(viii) 106480

Factors of 106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11

Hence, it’s not a perfect cube.

(ix) 166375

Factors of 166375 = 5 × 5 × 5 × 11 × 11 × 11 = 5³ × 11³ = 55³

Hence, it’s a perfect cube.

(x) 456533

Factors of 456533 = 11 × 11 × 11 × 7 × 7 × 7 = 11³ × 7³ = 77³

Hence, it’s a perfect cube.

i) 216 = 2³ × 3³ = 6³

It’s a cube of even natural number.

ii) 512 = 2⁹ = (2³)³ = 8³

It’s a cube of even natural number.

iii) 729 = 3³ × 3³ = 9³

It’s not a cube of even natural number.

iv) 1000 = 10³

It’s a cube of even natural number.

v) 3375 = 3³ × 5³ = 15³

It’s not a cube of even natural number.

vi) 13824 = 2² × 3⁴ × 41

Its not even a cube.

i) 125 = 5 × 5 × 5 × 5 = 5³

It’s a cube of odd natural number.

ii) 343 = 7 × 7 × 7 = 7³

It’s a cube of odd natural number.

iii) 1728 = 2⁶ × 3³ = 4³ × 3³ = 12³

It’s not a cube of odd natural number. As 12 is even number.

iv) 4096 = 2¹² = (2⁶)² = 64²

Its not even a cube.

v) 32768 = 2¹⁵ = (2⁵)³ = 32³

It’s a cube of odd natural number. As 32 is an even number.

vi) 6859 = 19 × 19 × 19 = 19³

It’s a cube of odd natural number.

(i) 675

Factors of 675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5²

Hence, to make a perfect cube we need to multiply the product by 5.

(ii) 1323

Factors of 1323 = 3 × 3 × 3 × 7 × 7 = 3³ × 7²

Hence, to make a perfect cube we need to multiply the product by 7.

(iii) 2560

Factors of 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 = 2³ × 2³ × 2³ × 5

Hence, to make a perfect cube we need to multiply the product by 5 × 5 = 25.

(iv) 7803

Factors of 7803 = 3 × 3 × 3 × 17 × 17 = 3³ × 17²

Hence, to make a perfect cube we need to multiply the product by 17.

(v) 107811

Factors of 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 3³ × 3 × 11³

Hence, to make a perfect cube we need to multiply the product by 3 × 3 = 9.

(vi) 35721

Factors of 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 3³ × 3³ × 7²

Hence, to make a perfect cube we need to multiply the product by 7.

(i) 675

Prime factors of 675 = 3 × 3 × 3 × 5 × 5 = 3³ × 5²

We find that 675 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 5² = 25, which gives 27 as quotient and we know that 27 is a perfect cube .

(ii) 8640

Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2³ × 2³ × 3³ × 5

We find that 8640 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 5 , which gives 1728 as quotient and we know that 1728 is a perfect cube.

(iii) 1600

Prime factors of 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2³ × 2³ × 5²

We find that 1600 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 5² = 25, which gives 64 as quotient and we know that 64 is a perfect cube

(iv) 8788

Prime factors of 8788 = 2 × 2 × 13 × 13 × 13 = 2² × 13³

We find that 8788 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube

(v) 7803

Prime factors of 7803 = 3 × 3 × 3 × 17 × 17 = 3³ × 17²

We find that 7803 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 17² = 289 , which gives 27 as quotient and we know that 27 is a perfect cube

(vi) 107811

Prime factors of 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11 = 3³ × 11³ × 3

We find that 107811 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube.

(vii) 35721

Prime factors of 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7 = 3³ × 3³ × 7²

We find that 35721 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 7² = 49, which gives 729 as quotient and we know that 729 is a perfect cube

(viii) 243000

Prime factors of 243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 2³ × 3³ × 5³ × 3²

We find that 243000 is not a perfect cube.

Hence, for making the quotient a perfect cube we divide it by 3² = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube

Let the number is = a

Cube of number will be = a³

Now, the number is trebled = 3 × a = 3a

Cube of new number = (3a)³ = 27a³

Hence, new cube is 27 times of original cube.
Hence, proved.

(i) 3?

Let the number is = a

Its cube will be = a³

According to the question, the number is multiplied by 3

New number become = 3a

Its new cube will be = (3a)³ = 27a³

Hence, cube will become

(ii) 4?

Let the number is = a

Its cube will be = a³

According to the question, the number is multiplied by 4

New number become = 4a

Its new cube will be = (4a)³ = 64a³

Hence, cube will become

(iii) 5?

Let the number is = a

Its cube will be = a³

According to the question, the number is multiplied by 5

New number become = 5a

Its new cube will be = (5a)³ = 125a³

Hence, cube will become

Area of one face of cube = 64 m² (Given)

Let length of edge edge of cube = a metre

=

=

Now, volume of cube = a³ =

Surface area of cube = 384 m² (Given)

Let length of each edge of cube = a metre

Volume of cube = a³ = (8)³ = 512m³

(i)

After solving we get,

(ii)

After solving we get,

i) 31

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 31 is = 1

Cube of 1 = 1³ = 1

Therefore, unit digit of cube of 31 is always be 1.

ii) 109

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 109 is = 9

Cube of 9 = 9³ = 729

Therefore, unit digit of cube of 109 is always be 9.

iii) 388

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 388 is = 8

Cube of 8 = 8³ = 512

Therefore, unit digit of cube of 388 is always be 2.

iv) 4276

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 4276 is = 6

Cube of 6 = 6³ = 216

Therefore, unit digit of cube of 4276 is always be 6.

v) 5922

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 5922 is = 2

Cube of 2 = 2³ = 8

Therefore, unit digit of cube of 5922 is always be 8.

vi) 77774

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 77774 is = 4

Cube of 4 = 4³ = 64

Therefore, unit digit of cube of 77774 is always be 4.

vii) 44447

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 44447 is = 7

Cube of 7 = 7³ = 343

Therefore, unit digit of cube of 44447 is always be 3.

viii) 125125125

To find unit digit of cube of a number we do the cube of unit digit only.

Here, unit digit of 125125125 is = 5

Cube of 5 = 5³ = 125

Therefore, unit digit of cube of 125125125 is always be 5.

(i) 35

we have , a = 3 and b = 5

Thus cube of 35 is 42875.

(ii) 56

we have , a = 5 and b = 6

Thus cube of 56 is 175616.

(iii) 72

we have , a = 7 and b = 2

Thus cube of 72 is 373248.

(i) 64

Prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ = 4³

Hence, it’s a perfect cube.

(ii) 216

Prime factors of 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³ = 6³

Hence, it’s a perfect cube.

(iii) 243

Prime factors of 243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3²

Hence, its not a perfect cube.

(iv) 1728

Prime factors of 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 3³ = 12³

Hence, it’s a perfect cube.

Only non-perfect cube in previous question was = 243

(a) Multiplied so that the product is a perfect cube.

Prime factors of 243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3²

Hence, to make it a perfect cube we should multiply it by 3.

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

Prime factors of 243 = 3 × 3 × 3 × 3 × 3 = 3³ × 3²

Hence, to make it a perfect cube we have to divide it by 9.

(i) If n is even, then n³ is also even.

Let the three even natural numbers be 2 , 4 , 6

Cubes of these numbers ,

= 2³ = 8

= 4³ = 64

= 6³ = 216

Hence, we can see that all cubes are even in nature.

Statement verified.

(ii) If n is odd, then n³ is also odd.

Let three odd natural numbers are = 3 , 5 , 7

Cubes of these numbers =

= 3³ = 27

= 5³ = 125

= 7³ = 343

Hence, we can see that all cubes are odd in nature.

Statement verified.

(iii) If n leaves remainder 1 when divided by 3, then n³ also leaves 1 as remainder when divided by 3.

Let three natural numbers of the form (3n+1) are = 4, 7 , 10

Cube of numbers = 4³ = 64 , 7³ = 343 , 10³ = 1000

We can see that if we divide these numbers by 3 , we get 1 as remainder in each case.

Statement verified.

(iv) If a natural number n is of the form 3p+2 then n³ also a number of the same type.

Let three natural numbers of the form (3p+2) are = 5 , 8 , 11

Cube of these numbers = 5³ = 125 , 8³ = 512 , 11³ = 1331

Now, we try to write these cubes in form of (3p + 2)

= 125 = 3 × 41 + 2

= 512 = 3 × 170 + 2

= 1331 = 3 × 443 + 2

Hence, statement verified.

(i) 392 is a perfect cube.

False.

Prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 2³ × 7²

(ii) 8640 is not a perfect cube.

True

Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2³ × 2³ × 3³ × 5

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

True

Beause a perfect cube always have zeros in multiple of 3.

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

False

64 is a perfect cube = 4 × 4 × 4 and it ends with 4.

(v) For an integer a, a³ is always greater than a².

False

In case of negative integers ,

=

(vi) If a and b are integers such that a²>b², then a³>b³.

False

In case of negative integers,

=

But ,

(vii) If a divides b, then a³ divides b³.

True

If a divides b =

=

For each value of b and a its true.

(viii) If a² ends in 9, then a³ ends in 7.

False

Let a = 7

7² = 49 and 7³ = 343

(ix) If a² ends in an even number of zeros, then a³ ends in 25.

False

Let a = 20

= a² = 20² = 400 and a³ = 8000

(x) If a² ends in an even number of zeros, then a³ ends in an odd number of zeros.

False

Let a = 100

= a² = 100² = 10000 and a³ = 100³ = 1000000

(i) -11

=

(ii) -12

=

(iii) -21

=

(i) -64

Prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ = 4³

We find that 64 is a perfect cube of negative integer – 4 .

(ii) -1056

Prime factors of 1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

We find that 1056 is not a perfect cube.

Hence, - 1056 is not a cube of negative integer

(iii) -2197

Prime factors of 2197 = 13 × 13 × 13 = 13³

We find that 2197 is a perfect cube.

Hence, - 2197 is a cube of negative integer – 13 .

(iv) -2744

Prime factors of 2744 = 2 × 2 × 2 × 7 × 7 × 7 = 2³ × 7³ = 14³

We find that 2744 is a perfect cube.

Hence, - 2744 is a cube of negative integer – 14 .

(v) -42875

Prime factors of 42875 = 5 × 5 × 5 × 7 × 7 × 7 = 5³ × 7³ = 35³

We find that 42875 is a perfect cube.

Hence, - 42875 is a cube of negative integer – 35.

(i) -5832

Prime factors of 5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 = 2³ × 3³ × 3³ = 18³

We find that 5832 is a perfect cube.

Hence, - 5832 is a cube of negative integer – 18 .

(ii) -2744000

Prime factors of 2744000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7 = 2³ × 2³ × 5³ × 7³

We find that 2744000 is a perfect cube.

Hence, - 2744000 is a cube of negative integer – 140 .

(i)

=

(ii)

=

(iii)

=

(iv)

=

(v)

=

(vi)

=

(vii) 0.3

=

(viii) 1.5

=

(ix) 0.08

=

(x) 2.1

=

(i)

We have,

=

Hence,

(ii)

We have,

=

Hence,

(iii) 0.001331

We have,

=

Hence, 0.001331 is a perfect cube of

(iv) 0.04

We have,

=

Hence, 0.04 is not a perfect cube.

(i) 64

We have,

64 – 1 = 63

63 – 7 = 56

56 – 19 =37

37 – 37 = 0

∵ Subtraction is performed 4 times.

Hence, cube root of 64 is 4.

(ii) 512

We have,

512 – 1 = 511

511 – 7 = 504

504 – 19 = 485

485 – 37 = 448

448 – 61 = 387

387 – 91 = 296

296 – 127 = 169

169 – 169 = 0

∵ Subtraction is performed 8 times.

Hence, cube root of 512 is 8.

(iii) 1728

We have,

1728 – 1 = 1727

1727 – 7 = 1720

1720 – 19 = 1701

1701 – 37 = 1664

1664 – 91 = 1512

1512 – 127 = 1385

1385 – 169 = 1216

1216 – 217 = 999

999 – 271 = 728

728 – 331 = 397

397 – 397 = 0

∵ Subtraction is performed 12 times.

Hence, cube root of 1728 is 12.

(i) 130

Applying subtraction method, We have,

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

∵ Next number to be subtracted is 91, which is greter than 5

Hence, 130 is not a perfect cube.

(ii) 345

Applying subtraction method, We have,

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

∵ Next number to be subtracted is 169, which is greter than 2

Hence, 345 is not a perfect cube

(iii) 792

Applying subtraction method, We have,

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

∵ Next number to be subtracted is 271, which is greter than 63

Hence, 792 is not a perfect cube

(iv) 1331

Applying subtraction method, We have,

1331 – 1 = 1330

1330 – 7 = 1323

1323 – 19 = 1304

1304 – 37 = 1267

1267 – 61 = 1206

1206 – 91 = 1115

1115 – 127 = 988

988 – 169 = 819

819 – 217 = 602

602 – 271 = 331

331 – 331 = 0

∵ Subtraction is performed 11 times.

Hence, 1331 is a perfect cube

In previous question there are three numbers which are not perfect cubes.

i) 130

Apply subtraction method,

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

∵ Next number to be subtracted is 91, which is greter than 5

Hence, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from it.

130 – 5 = 125 (which is a perfect cube of 5)

ii) 345

Apply subtraction method,

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

∵ Next number to be subtracted is 169, which is greter than 2

Hence, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from it.

345 – 2 = 343 (which is a perfect cube of 7)

iii) 792

Apply subtraction method,

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

∵ Next number to be subtracted is 271, which is greter than 63

Hence, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from it.

792 – 63 = 729 (which is a perfect cube of 9)

(i) 343

By prime factorization method,

= =

(ii) 2744

By prime factorization method,

=

(iii) 4913

By prime factorization method,

=

(iv) 1728

By prime factorization method,

=

(v) 35937

By prime factorization method,

=

(vi) 17576

By prime factorization method,

=

(vii) 134217728

By prime factorization method,

=

(viii) 48228544

By prime factorization method,

=

= 2 × 2 × 7 × 13 = 364.

(ix) 74088000

By prime factorization method,

=

= 2 × 2 × 3 × 5 × 7 = 420.

(x) 157464

By prime factorization method,

=

= 2 × 3 × 3 × 3 = 54.

(xi) 1157625

By prime factorization method,

=

(xii) 33698267

By prime factorization method,

=

First we find out the prime factors of 3600,

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 2³ × 3² × 5² × 2

∵ only one triples is formed and three factors remained ungrouped in triples.

Hence, 3600 is not a perfect cube.

To make it a perfect cube we have to multiply it by ( 2 × 2 × 3 × 5) = 60

3600 × 60 = 216000 ( which is a perfect cube of 60)

First we find out the prime factors of 210125,

210125 = 5 × 5 × 5 × 41 × 41

∵ one triples remained incomplete, hence 210125 is not a perfect cube.

We see that if we multiply the factors by 41, we will get 2 triples as 2³ and 41³.

And the product become:

210125 × 41 = 8615125 = 5 × 5 × 5 × 41 × 41 × 41

Cube root of product =

First we find out prime factors of 8192,

8192 =

∵ one triples remain incomplete, hence 8192 is not a perfect cube.

So, we divide 8192 by 2 to make its quotient a perfect cube.

Cube root of 4096 =

Let the numbers are = x, 2x and 3x

According to the question,

So, the numbers are ,

x = 14

2x = 2 × 14 = 28

Volume of cube = 9261000 m³

Let the side of cube = a metre

So,

=

=

=

Hence, the side of cube = 210 metre

(i) We have,

Cube root of -125 =

(ii) We have,

Cube root of -5832 =

So to find out the cube root of 5832, we will use the mehod of unit digits.

Let’s take number 5832.

Unit digit = 2

So unit digit in the cube root of 5832 = 8

After striking out the units, tens and hundreds digits of 5832,

Now we left with 5 only.

As we know that 1 is the Largest number whose cube is less than or equals to 5.

So,

The tens digit of the cube root of 5832 is 1.

(iii) We have,

We will use the method of factorization to find out the cube root of 2744000

Factorizing 2744000 into prime factors,

We get,

2744000 = 2×2×2×2×2×2×5×5×5×7×7×7

Now group the factors into triples of equal factors, we get,

2744000 = (2×2×2) ×(2×2×2) ×(5×5×5) ×(7×7×7)

As we can see that all the prime factors of 2744000 can be grouped in to triples of equal factors and no factor is left over.

Now take one factor from each group and by multiplying we get,

2×2×5×7 = 140

So we can say that 2744000 is a cube of 140

Hence,

(iv) We have,

By using unit digit method,

Let’s take Number = 753571

Unit digit = 1

So unit digit in the cube root of 753571 = 1

After striking out the units, tens and hundreds digits of 753571,

Now we left with 753.

As we know that 9 is the Largest number whose cube is less than or equals to 753(9³<753<10³).

So,

The tens digit of the cube root of 753571 is 9.

(v) We have,

By using unit digit method, we will find out the cube root of 32768,

Let’s take Number = 32768

Unit digit = 8

So unit digit in the cube root of 32768 = 2

After striking out the units, tens and hundreds digits of 32768,

Now we left with 32.

As we know that 9 is the Largest number whose cube is less than or equals to 32(3³<32<4³).

So,

The tens digit of the cube root of 32768 is 3.

(i) Given,

As we know LHS = RHS, so the equation is true.

(ii) Given,

LHS = RHS

(iii) Given,

LHS = RHS

(iv) Given,

LHS =

RHS =

LHS = RHS

(i) We know that for any two integers a and b,

So from this property, we have:

(ii) By Applying a and b, , we have

To find out cube root by using units digit:

Let’s take the number 1728.

So,

Unit digit = 8

The unit digit in the cube root of 1728 = 2

After striking out the units, tens and hundreds digits of the given number, we are left with the 1.

As we know 1 is the largest number whose cube is less than or equals to 1.

So,

The tens digit of the cube root of 1728 = 1

Prime factors of 216 = 2×2×2×3×3×3

On grouping the factors in triples of equal factor,

We have,

216 = {2×2×2}×{3×3×3}

Taking one factor from each group we get,

So,

(iii) By Applying a and b propertise, , we have

To find out cube root by using units digit:

Let’s take the number 2744.

So,

Unit digit = 4

The unit digit in the cube root of 2744= 4

After striking out the units, tens and hundreds digits of the given number, we are left with the 2.

As we know 1 is the largest number whose cube is less than or equals to 2.

So,

The tens digit of the cube root of 2744 = 1

Prime factors of 216 = 2×2×2×3×3×3

On grouping the factors in triples of equal factor,

We have,

216 = {2×2×2}×{3×3×3}

Taking one factor from each group we get,

So,

(iv) By Applying a and b properties,, we have

To find out cube root by using units digit:

Let’s take the number 15625.

So,

Unit digit = 5

The unit digit in the cube root of 15625 = 5

After striking out the units, tens and hundreds digits of the given number, we are left with the 15.

As we know 2 is the largest number whose cube is less than or equals to 15(2³<15<3³).

So,

The tens digit of the cube root of 15625 = 2

Also

As we know 9×9×9 = 729

Thus,

(i)

We have,

=

(ii)

_{We have,}

₌

(iii)

_{We have,}

=

Getting prime factors of numbers,

=

= 2 × 5 × 7 = 70.

(iv)

_{We have,}

=

=

(i)

_{We have,}

= =

(ii)

By getting prime factors of given problems. We have,

=

(iii)

By getting prime factors of given problems. We have,

=

(iv)

By getting prime factors of given problems. We have,

=

(v)

By getting prime factors of given problems. We have,

=

(i) 0.001728

Given,

0.001728 =

Getting prime factors of 1728,

1728 = 2×2×2×2×2×2×3×3×3 =

,

(ii) 0.003375

Given,

0.003375 =

Getting prime factors of 3375 ,

3375 = 3×3×3×5×5×5,

Also,

(iii) 0.001

Given,

∵

(iv) 1.331

Given

∵ 1.331 =

(i)

_{By prime factorization of terms, We have,}

₌₌ =

=

(ii)

_{By prime factorization of terms, We have,}

=

=

(iii)

_{By prime factorization of terms, We have,}

=

(iv)

_{By prime factorization of terms, We have,}

= =

=

(v)

_{By prime factorization of terms, We have,}

=

(i)

We have,

LHS =

RHS =

∵ LHS = RHS

Hence, equation is true.

(ii)

We have,

LHS =

RHS =

∵ LHS = RHS

Hence, equation is true.

(i)

_{We have,}

=

Hence,

= 5

(ii)

_{We have,}

=

Hence,

=

(iii)

We have,

=

= 4 × 3

Hence,

= 3

(iv)

_{We have,}

= =

=

Hence,

=

(v)

_{We have,}

=

=

Hence,

=

(vi)

_{We have,}

=

=

Hence,

=

(vii)

We have,

=

Hence,

=

(viii)

We have,

=

Hence,

=

(ix)

= 8

Given,

Volume of a cube = 474.552 cubic metres

V = 8³,

S = side of the cube

So,

8³ = 474.552 cubic metres

= 8 =

On factorising 474552 into prime factors, we get:

474552 = 2×2×2×3×3×3×13×13×13

On grouping the factors in triples of equal factors, we get:

474552 = {2×2×2}×{3×3×3}×{13×13×13}

Now taking 1 factor from each group we get,

Also,

So, length of the side is 7.8m.

Lest assume the numbers be 2a, 3a, and 4a.

According to the question:

(2a)³+ (3a)³+(4a)³ = 0.334125

= 8a³+27 a³+64 a³ = 0.334125

=99 a³ = 0.334125

= a³ =

= a =

= a =

= a =

Thus the numbers are:

2×0.15 = 0.30

3×0.15 = 0.45

4×0.15 = 0.60

Given,

Volume of the side s =

(i)

_{We have,}

=

Now by prime factorization method,

=

=

(ii)

_{We have,}

= =

Now by prime factorization method,

=

=

(iii)

_{We have,}

=

Now by prime factorization method,

=

=

(iv)

_{We have,}

=

Now by prime factorization method,

=

(i)

_{Taking cube root of the whole, we get,}

=

We know that,

=

=

Now by prime factorization,

=

(ii)

Taking cube root of the whole,

=

We know that,

=

=

Now by prime factorization,

=

= 3 × 7 × 13 = 273.

(iii)

Taking cube root of the whole,

=

We know that,

=

=

Now by prime factorization,

=

= 5 × 7 × 17 = 595.

(iv)

Taking cube root of the whole, we get,

=

We know that,

=

=

Now by prime factorization method,

=

= 5 × 7 ×11 = 385.

(i) 226981

Let’s consider the number 226981.

Unit digit = 1

The unit digit of the cube root of 226981 = 1

(ii) 13824

Let’s consider the number 13824.

Unit digit = 4

The unit digit of the cube root of 13824 = 4

(iii) 571787

Let’s consider the number 571787.

Unit digit = 7

The unit digit of the cube root of 571787 = 3

(iv) 175616

Let’s consider the number 175616.

Unit digit = 6

The unit digit of the cube root of 175616 = 6

(i) 226981

Let’s take number 226981.

Unit digit = 1

So unit digit in the cube root of 226981 = 1

After striking out the units, tens and hundreds digits of 226981,

Now we left with 226 only.

As we know that 6 is the Largest number whose cube root is less than or equals to 226(6³<226<7³).

So,

The tens digit of the cube root of 226981 is 6.

(ii) 13824

Let’s take number 13824.

Unit digit = 4

So unit digit in the cube root of 13824 = 4

After striking out the units, tens and hundreds digits of 13824,

Now we left with 13 only.

As we know that 2 is the Largest number whose cube root is less than or equals to 13(2³<13<3³).

So,

The tens digit of the cube root of 13824 is 2.

(iii) 571787

Let’s take number 571787.

Unit digit = 7

So unit digit in the cube root of 571787 = 3

After striking out the units, tens and hundreds digits of 571787,

Now we left with 571 only.

As we know that 8 is the Largest number whose cube root is less than or equals to 571(8³<571<9³).

So,

The tens digit of the cube root of 571787 is 8.

(iv) 175616

Let’s take number 175616.

Unit digit = 6

So unit digit in the cube root of 175616 = 6

After striking out the units, tens and hundreds digits of 175616,

Now we left with 175 only.

As we know that 5 is the Largest number whose cube root is less than or equals to 175(5³<175<6³).

So,

The tens digit of the cube root of 175616 is 5.

As we know that 7 lies between 1 and 100 so by using cube root table we have,

So, Answer is 1.913.

As we know that 70 lies between 1 and 100 so by using cube root table from column x

we have,

So, Answer is 4.121

Given,

700 = 70×10

By using cube root table 700 will be in the column against 70.

So we have,

7000 = 70×100

By using cube root table,

We get,

1100 = 11×100

By using cube root table,

We get,

780 = 78×10

By using cube root table 780 would be in column against 78.

So we get,

7800 = 78×100

By using cube root table,

We get,

By primefactorisation method,

We get,

1346 = 2×673

Also,

670<673<680 =>

By using cube root table,

For the difference (680-670) which is 10.

The difference in the values,

= 8.794 - 8.750 = 0.044

For the difference (673-670) which is 3.

The difference in the values,

=

So,

250 = 25×100

By using cube root table 250 would be in column against 25.

So we get,

=

From cube root table we get,

=

Hence,

=

Thus, the required cube root is = 17.227.

= =

From cube root table we get,

=

Hence,

=

Thus, the required cube root is = 21.40.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (740 – 730 = 10 ) difference in cube root values = 9.045 – 9.004 = 0.041

∴ For difference (732 – 730 = 2) difference in cube root values =

=

Thus, the required cube root is = = 9.012.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (7400 – 7300 = 100 ) difference in cube root values = 19.48 – 19.39 = 0.09

∴ For difference (7342 – 7300 = 42) difference in cube root values =

=

Thus, the required cube root is = 19.427.

=

From cube root table we get,

=

Hence,

=

Thus, the required cube root is = 51.062.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (180 – 170 = 10 ) difference in cube root values = 5.646 – 5.540 = 0.106

∴ For difference (175 – 170 = 5) difference in cube root values =

=

Hence,

=

Thus, the required cube root is 33.558.

=

From cube root table we get,

=

Hence,

=

Thus the required cube root is = 0.646.

=

From cube root table we get,

=

Hence,

=

Thus the required cube root is = 2.049.

=

From cube root table we get,

=

Hence,

=

Thus the required cube root is = 0.951.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (870 – 860 = 10 ) difference in cube root values = 9.546 – 9.510 = 0.036

∴ For difference (865 – 860 = 5) difference in cube root values =

=

We also have,

∴

Thus the required cube root is = 2.053.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (7600 – 7500 = 100 ) difference in cube root values = 19.66 – 19.57 = 0.09

∴ For difference (7532 – 7500 = 32) difference in cube root values =

=

Thus the required cube root is = 19.599.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (840 – 830 = 10 ) difference in cube root values = 9.435 – 9.398 = 0.037

∴ For difference (833 – 830 = 3) difference in cube root values =

= = 9.409

Thus the required cube root is = 9.409.

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (350 – 340 = 10 ) difference in cube root values = 7.047 – 6.980 = 0.067

∴ For difference (342 – 340 = 2) difference in cube root values =

=

We also have,

∴

Thus the required cube root is = 3.246.

Volume of cube = 275 cm³ (Given)

Let side of cube = a cm

So,

=

=

We know that value of

From cube root table we get,

=

So by unitary method,

∵ For difference (280 – 270 = 10 ) difference in cube root values = 6.542 – 6.463 = 0.079

∴ For difference (275 – 270 = 5) difference in cube root values =

Hence,

Thus the required cube root is = 6.503.