Which of the following numbers are perfect cubes?

(i) 125

Resolving 125 into prime factors:

$125=5 \times 5 \times 5$

Here, one triplet is formed, which is $5^{3}$. Hence, 125 can be expressed as the product of the triplets of 5 .

Therefore, 125 is a perfect cube.

(ii) 243 is not a perfect cube.

(iii) 343

Resolving 125 into prime factors:

$343=7 \times 7 \times 7$

Here, one triplet is formed, which is $7^{3}$. Hence, 343 can be expressed as the product of the triplets of $7 .$

Therefore, 343 is a perfect cube.

(iv) 256 is not a perfect cube.

(v) 8000

Resolving 8000 into prime factors:

$8000=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5$

Here, three triplets are formed, which are $2^{3}, 2^{3}$ and $5^{3}$. Hence, 8000 can be expressed as the product of the triplets of 2,2 and 5, i.e. $2^{3} \times 2^{3} \times 5^{3}=20^{3}$.

Therefore, 8000 is a perfect cube.

(vi) 9261

Resolving 9261 into prime factors:

$9261=3 \times 3 \times 3 \times 7 \times 7 \times 7$

Here, two triplets are formed, which are $3^{3}$ and $7^{3}$. Hence, 9261 can be expressed as the product of the triplets of 3 and 7, i.e. $3^{3} \times 7^{3}=21^{3}$.

Therefore, 9261 is a perfect cube.

(vii) 5324 is not a perfect cube.

(viii) 3375 .

Resolving 3375 into prime factors:

$3375=3 \times 3 \times 3 \times 5 \times 5 \times 5$

Here, two triplets are formed, which are $3^{3}$ and $5^{3}$. Hence, 3375 can be expressed as the product of the triplets of 3 and 5, i.e. $3^{3} \times 5^{3}=15^{3}$

Therefore, 3375 is a perfect cube.