Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube. (i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704
(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, Â [81]Â 3= [3 x 3 x 3 x 3]Â 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, Â [128]Â 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 = Â 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]Â 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, Â Â [192]Â 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, Â 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.
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