chapter -7 Cubes and Cube Roots

Explanation:

By doing prime factorisation of 216,

We can say, 216 = 2×2×2×3×3×3

216 = (2×2×2)×(3×3×3)

Here, , 216 can be separated into group of 3 which are of equal factors, i.e.,

∴ 216 = (2×3) = 6

Hence, 216 is the cube of 6.

Explanation:

By doing prime factorisation of 128

128 = 2×2×2×2×2×2×2

By grouping the factors in triplets of equal factors, 128 = (2×2×2)×(2×2×2)×2

Here, 128 cannot be seperated into group of 3 of equal factors, as we are left with one factor: 2.

Therefore, 128 is not a perfect cube.

Explanation:

By doing prime factorisation of 1000

1000 = 2×2×2×5×5×5

Here, , 1000 can be separated into group of 3 which are of equal factors, i.e.,

, (2×2×2)×(5×5×5)

∴ 1000 = (2×5) =10

Therefore, 1000 is the cube of 10.

Explanation:

By doing prime factorisation of 1000,

100 = 2×2×5×5

Here, 100 cannot be separated into group of 3 of equal factors,

Thus, 100 is not a perfect cube.

Explanation:

By doing prime factorisation of 46656,

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

Here, 46656 can be separated into group of 3 which are of equal factors, i.e.,

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

∴ 46656 = (2×2×3×3) = 36

Therefore, 46656 is the cube of 36.

Explanation:

By doing prime factorisation of 243

243 = 3×3×3×3×3

243 = (3×3×3)×3×3

243 cannot be separated into group of 3 which are of equal factors, i.e.,

Therefore, we will multiply 243 by 3 to get the perfect cube.

Explanation:

By taking LCM of 256

256 = 2×2×2×2×2×2×2×2

256 = (2×2×2)×(2×2×2)×2×2

Here, 2 cannot be grouped into triplets of equal factors.

Therefore, we will multiply 256 by 2 to get the perfect cube.

Explanation:

By taking LCM of 72,

72 = 2×2×2×3×3

72 = (2×2×2)×3×3

Here, 3 cannot be grouped into triplets of equal factors.

Therefore, we will multiply 72 by 3 to get the perfect cube.

Explanation:

By taking LCM of 675

675 = 3×3×3×5×5

675 = (3×3×3)×5×5

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

Thus, we will multiply 675 by 5 to get the perfect cube.

Explanation:

By taking LCM of 100,

100 = 2×2×5×5

Here, 2 and 5 cannot be grouped into triplets of equal factors.

Therefore, we will multiply 100 by (2×5) 10 to get the perfect cube.

Explanation:

By taking LCM of 81,

81 = 3×3×3×3

81 = (3×3×3)×3

Here, 3 cannot be separated into group of 3 of equal factors.

Therefore, we will divide 81 by 3 to get the perfect cube.

Explanation:

By taking LCM of 128,

128 = 2×2×2×2×2×2×2

128 = (2×2×2)×(2×2×2)×2

Here, 2 cannot be separated into group of 3 of equal factors.

Hence, will divide 128 by 2 to get the perfect cube.

Explanation:

By taking prime factorisation of 135,

135 = 3×3×3×5

135 = (3×3×3)×5

Here, 5 is left out whereas 3 can divided into group of 3

Therefore, we will divide 135 by 5 to get the perfect cube.

Explanation:

By taking prime factorisation of 192,

192 = 2×2×2×2×2×2×3

192 = (2×2×2)×(2×2×2)×3

Here, 3 cannot be grouped into triplets of equal factors.

Therefore, we will divide 192 by 3 to get the perfect cube.

Explanation:

By taking prime factorisation of 704,

704 = (2×2×2)×(2×2×2)×11

Here, 2 is divided into group of three twice whereas 11 cannot be grouped into triplets of equal factors.

Hence, we will divide 704 by 11 to get the perfect cube.

Explanation:

64 = 2×2×2×2×2×2

Here, 64 can be separated into group of 3 which are equal factors,

64 = (23)×(23)

∴ 364 = 2×2 = 4

Hence, cube root of 64 is 4.

Explanation:

512 = 2×2×2×2×2×2×2×2×2

Here, 512 can be separated into group of 3 thrice which are equal factors,

∴ 3512 = 2×2×2 = 8

Hence, 512 is cube root of 8

Explanation:

10648 = 2×2×2×11×11×11

Here, 10648 can be separated into group of 3 with multiple of 2 and 11 which are equal factors,

∴ 310648 = 2 ×11 = 22

Hence, cube root of 10648 is 22

Explanation:

27000 = 2×2×2×3×3×3×5×5×5

 Here, 27000 can be separated into group of 3 which includes multiple of 2, 3 and 5 which are equal factors,

∴ 327000= (2×3×5) = 30

Hence, cube root of 30 is 27000

Explanation:

15625 = 5×5×5×5×5×5

Here, 15625 can be separated into group of 3 of multiple 5 twice which are equal factors,

∴ 315625 = (5×5) = 25

Hence, cube root of 15625 is 25

Explanation:

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

Here, 13824 can be separated into group of 3 of multiple 2 thrice and 3 once which are equal factors,

∴ 13824 = (2×2× 2×3) = 24

Hence, 24 is the cube root of 13824.

Explanation:

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 110592 can be grouped into triplets of equal factors.

∴ 110592 = (2×2×2×2 × 3) = 48

Hence, 48 is the cube root of 110592.

Explanation:

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triplets of equal factors.

∴ 46656 = (2×2×3×3) = 36

Hence the cube root of 46656 is 36

Explanation:

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)

Here, 175616 can be grouped into group of 3 of equal factors.

∴ 175616 = (2×2×2×7) = 56

Hence, the cube root of 175616 is 56

Explanation:

91125 = 3×3×3×3×3×3×3×5×5×5

91125 = (3×3×3)×(3×3×3)×(5×5×5)

Here, 91125 can be grouped into triplets of equal factors.

∴ 91125 = (3×3×5) = 45

Hence, the cube root of 91125 is 45

Explanation:

False, cube of an odd number is always odd.

Explanation:

True

Explanation:

False, it is not necessary that always cube of 5 ends with 25, for example, the cube of 15 is 3375 which does not end with 25.

Explanation:

False, this can be justified with the help of this example, 123= 12 x 12 x 12= 1728, which ends with 8 and is a perfect cube.

Explanation:

False, cube of two digit number can never be a three digit number.

The smallest two digit number is 10 and its cube root is 1000.

Explanation:

False

The largest two digit number is 99 and its cube root is 970299 which is 6 digits.

Explanation:

True as the cube of 1 is 1 and 2 is 8.

Explanation:

i) Let us first separate the digits in 1331 as 1 and 331

Now, the digit in the unit place is 1

We know, the cube of 1 is always 1

Thus, the unit digit of cube root of 1331 is 1

Now, considering the other group, 1

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

Therefore, the ten’s digit of our cube root is taken as the unit place of the smallest number.

We know that the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.

∴³ _/1331 = 11

(ii) Let us first separate the digits in 4913 as 4 and 913

The digit in unit place is 3 

We know that cube of 7 ends with 3. So, we consider the unit digit of cube root as 7

Now, considering the other group, 4

We know 13 = 1 and 23 = 8 , 1 < 4 < 8

Thus, 1 is taken as the tens digit of the cube root.

∴ ³_/4913= 17

(iii) Let us first separate the digits in 12167 as 12 and 167

The digit in unit place is 7

We know, cube of 3 ends with 7. So, we consider the digit in unit place of cube root as 3

Now considering the other group, 12

We know 23 = 8 and 33 = 27 , 8 <12 <27

Thus, 2 is taken as the tens digit of the cube root.

∴ ³_/12167= 23

(iv) Let us first separate the digits in 32768 as 32 and 768.

The digit in unit place is 8. We know that the cube of 2 is 8. So, we consider the unit place of cube root as 2.

Considering, the other group 32

We know 33 = 27 and 43 = 64 , 27 <32 < 64

Thus, 3 is taken as the tens digit of the cube root.

∴ ³_/32768= 32