(i) 125

First find out the prime factors of 125,

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:

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:

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:

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:

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:

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:

5324 = 2×2×11×11×11

Therefore, 5324 is not a perfect cube.

(viii) 3375

The prime factorization of 3375 is shown below:

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.