Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
(i) 125 (ii) 243
(iii) 343 (iv) 256
(v) 8000 (vi) 9261
(vii) 324 (viii) 3375
(i) 125
First find out the prime factors of 125,
5 | 125 |
5 | 25 |
5 | 5 |
1 |
125 = 5×5×5
As we see a group of three 5 is made, which we can also be write as 53;
So, 125 is the product of triplets of 5.
Therefore, it is the perfect cube.
(ii) 243
The prime factorization of 256 is shown below:
3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
1 |
243 = 3 × 3 × 3 × 3 × 3
To be a perfect cube the prime factors of number should make a group of 3 but as we can see here more than 3 numbers are available in prime factors.
So, 243 is not the perfect cube.
(iii) 343
The prime factorization of 256 is shown below:
7 | 343 |
7 | 49 |
7 | 7 |
1 |
343 = 7 × 7 × 7
As we see a group of three 7 is formed, which we can also be write as 73;
So, 343 is the product of triplets of 7.
Therefore, it is the perfect cube.
(iv) 256
The prime factorization of 256 is shown below:
2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
1 |
256 = 2×2×2×2×2×2×2×2
If the prime factors are not making the pairs of three so the number is not perfect cube.
(v) 8000
The prime factorization of 8000 is shown below:
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
As we can see three pairs can be made of the above prime factors, which are 23, 23, and 53.
So, 8000 can be expressed as the product of the triplets of 2, 2 and 5, i.e.
23 × 23 × 53 = 203
Therefore, 8000 is a perfect cube.
(vi) 9261
The prime factorization of 9261 is shown below:
3 | 9261 |
3 | 3087 |
3 | 1029 |
7 | 343 |
7 | 49 |
7 | 7 |
1 |
9261 = 3×3×3×7×7×7
As we can see two pairs can be made of the above prime factors, which are 33, and 73.
So, 9261 can be expressed as the product of the triplets of 3 and 7, i.e.
33 × 73 = 213
Therefore, 9261 is a perfect cube.
(vii) 5324
The prime factorization of 5324 is shown below:
2 | 5324 |
2 | 2662 |
11 | 1331 |
11 | 121 |
11 | 11 |
1 |
5324 = 2×2×11×11×11
Therefore, 5324 is not a perfect cube.
(viii) 3375
The prime factorization of 3375 is shown below:
3 | 3375 |
3 | 1125 |
3 | 375 |
5 | 125 |
5 | 25 |
5 | 5 |
1 |
3375 = 3×3×3×5×5×5
As we can see two pairs can be made of the above prime factors, which are 33, and 53.
So, 3375 can be expressed as the product of the triplets of 3 and 5, i.e.
33 × 53 = 153
Therefore, 3375 is a perfect cube.