\(\sqrt [ 3 ]{ \cfrac { 9261 }{ 8000 } } =\cfrac { \sqrt [ 3 ]{ 9261 } }{ \sqrt [ 3 ]{ 8000 } } =\cfrac { \left( 21\times 21\times 21 \right) ^{ 1/3 } }{ \left( 20\times 20\times 20 \right) ^{ 1/3 } } =\cfrac { 21 }{ 20 } \)

By prime factorisation, we have 27000 = 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5
\(\therefore \sqrt [ 3 ]{ 2700 } =2\times 3\times 5=30\)

By prime factorization, we have \(400=\underbrace { 2\times 2\times 2 } \times 2\times 5\times \)
There is only one triplet. To make further triplets, we will need two more 2’s and one more 5.
Therefore, 400 is not a perfect cube.

\(\sqrt { 256 } =\sqrt { 16\times 16 } =\sqrt { 16 } \times \sqrt { 16 } =4\times 4=16\) or \(\sqrt { 256 } =\sqrt { 64\times 4 } =\sqrt { 64 } \times \sqrt { 4 } =8\times 2=16\)
Try to fill up the following table of similar problems using \(\sqrt { a\times b } =\sqrt { a } \times \sqrt { b } \)

We find 250 = 5 x 5 x 5 x 2
= 5² x 5 x 2
Here, the prime factors 5 and 2 do not have pairs.
Therefore, we can either divide 250 by 10 (5 x 2) or multiply 250 by 10.
(i) If we multiply 250 by 10, we get 2500 = 5² × 5 × 2 × 5 × 2 and therefore the square root for 2500 would be 5 x 5 x 2 = 50.
(ii) If we divide 250 by 10 we get 25 and in that case we get \(\sqrt { 25 } =\sqrt { { 5 }^{ 2 } } =5\)