Given : Is 0 a perfect cube. We know that 0 is not a natural number. Hence, it is not a perfect cube number.

Find the cube roots of following number 223 678 3

Given, 223 678 3 Cube root of 223 678 3 is = 223 678 3 3 = 223 678 × 223 678 × 223 678 3 = 223 678

We need to find, the cube root of - 1331 Let us find the prime factors of 1331 11 1331 11 121 11 11 1 ∴ 1331 = 11 × 11 × 11 ⇒ - 1331 = - 11 × 11 × 11 Then, to obtain the cube root, choosing one number from the triplet ∴ - 1331 3 = - 11

We need to find, cube root of 9261 Let us find the prime factors of 9261 , we get, 3 9261 3 3087 3 1029 7 343 7 49 7 7 1 ∴ 9261 = 3 × 3 × 3 × 7 × 7 × 7 Now, making the group of triplets, ⇒ 9261 = ( 3 × 3 × 3 ) × ( 7 × 7 × 7 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 9261 3 = 3 × 7 = 21

We need to find, the cube root of - 216 Let us find the prime factors of 216 , we get, 2 216 2 108 2 54 3 27 3 9 3 3 1 ∴ 216 = 2 × 2 × 2 × 3 × 3 × 3 Now, making the group of triplets, ⇒ 216 = ( 2 × 2 × 2 ) × ( 3 × 3 × 3 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ - 216 3 = - ( 2 × 3 ) = - 6

We need to find, cube root of 8000 Let us find the prime factors of 8000 , we get, 2 8000 2 4000 2 2000 2 1000 2 500 2 250 5 125 5 25 5 5 1 ∴ 8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 Now, making the group of triplets, ⇒ 8000 = ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 5 × 5 × 5 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 8000 3 = 2 × 2 × 5 = 20

We need to find, cube root of 1728 Let us find the prime factors of 1728 , we get, 2 1728 2 864 2 432 2 216 2 108 2 54 3 27 3 9 3 3 1 ∴ 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 Now, making the group of triplets, ⇒ 1728 = ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 3 × 3 × 3 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 1728 3 = ( 2 × 2 × 3 ) = 12

We need to find, cube root of 3375 Let us find the prime factors of 3375 , we get, 5 3375 5 675 5 135 3 27 3 9 3 3 1 ∴ 3375 = 3 × 3 × 3 × 5 × 5 × 5 Now, making the group of triplets, ⇒ 3375 = ( 3 × 3 × 3 ) × ( 5 × 5 × 5 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 3375 3 = 3 × 5 = 15

We need to find, cube root of 4096 Let us find the prime factors of 4096 , we get, 2 4096 2 2048 2 1024 2 512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1 ∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Now, making the group of triplets, ⇒ 4096 = ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 4096 3 = 2 × 2 × 2 × 2 = 16

Evaluate: 1331 3 1728 3

Given number is 1331 3 1728 3 Evaluating using the properties of cube roots. 1331 3 1728 3 = 11 × 11 × 11 3 ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 3 × 3 × 3 ) 3 = 11 2 × 2 × 3 = 11 12 . Hence, the result is 11 12 .

We need to find, the cube of 4096 Now, let us find the prime factors of 4096 , we get, 2 4096 2 2048 2 1024 2 512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1 ∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Taking a cube on both the sides, we get, 4096 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 3 ⇒ 4096 3 = 2 × 2 × 2 × 2 ⇒ 4096 3 = 16

We need to find, the value of 343 3 Let us find the prime factors of 343 , we get, 7 343 7 49 7 7 1 ∴ 343 = 7 × 7 × 7 Then, to obtain the cube root, let us find the product of primes, choosing one from the triplet ∴ 343 3 = 7

We need to find, cube root of - 512 Let us find the prime factors of - 512 , we get, 2 512 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1 ∴ 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 Now, making the group of triplets, ⇒ 512 = ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ - 512 3 = - 2 × 2 × 2 = - 8

We need to find, the cube root of 729 Let us find the prime factors of 729 , we get, 3 729 3 243 3 81 3 27 3 9 3 3 1 729 = 3 × 3 × 3 × 3 × 3 × 3 Now, making the group of triplets, ⇒ 729 = 3 × 3 × 3 × 3 × 3 × 3 Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 729 3 = 3 × 3 = 9

We need to find, the value of 64 3 Let us find the prime factors of 64 , we get, 2 64 2 32 2 16 2 8 2 4 2 2 1 ∴ 64 = 2 × 2 × 2 × 2 × 2 × 2 Now, making the group of triplets, ⇒ 64 = ( 2 × 2 × 2 ) × ( 2 × 2 × 2 ) Then, to obtain the cube root, let us find the product of primes, choosing one from each triplet ∴ 64 3 = 2 × 2 = 4