(i) We have 81 = 3 x 3 x 3 x 3


Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is  not a perfect cube
Now,   [81] 3= [3 x 3 x 3 x 3] 3
or    27 = 3 x 3 x 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

Grouping the prime factors of 128 into triples, we are left with 2
∴  128 is  not a perfect cube
Now,  [128] 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2]2
or         64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
∴  the smallest required number is 2.

(iii) we have 135 =  3 x 3 x 3 x 5

Grouping the prime factors of 135 into triples, we are left over with 5.
∴  135 is not a perfect cube
Now, [135]5 = [ 3 x 3 x 3 x 5] 5
or      27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

Grouping the prime factors of 192 into triples, 3 is left over.
∴  192 is not a perfect cube.
Now,     [192] 3= [2 x 2 x 2 x 2 x 2 x 2 x 3]3
or            64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube.
Thus,  the required smallest number is 3.
(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

Grouping the prime factors of 704 into triples, 11 is left over
∴  [704]11 = [2 x 2 x 2 x 2 x 2 x 2 x 11]11
or    64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
Thus, the required smallest number is 11.